MCQ Questions for Class 11 Engineering Maths Binomial Theorem Quiz 10

If the second term of the expansion

${ [{ a }^{ { 1 }/{ 13 } }+\frac { a }{ \sqrt { { a }^{ -1 } } } ] }^{ n }$ is

$14{ a }^{ { 5 }/{ 2 } }$ , then the value of

$\frac { ^{ n }{ C }_{ 3 } }{^{ n } { C }_{ 2 } }$ is .

0%
4
0%
3
0%
12
0%
6

In the expansion of

$\left( 1+x \right) ^{ n }$ , The binomial coefficients of three consecutive terms are respectively 220, 495 and 795, the value

0%
$10$
0%
$11$
0%
$12$
0%
$13$

Explanation

Suppose the three consecutive terms are

$T_{r-1},\,T_r$ and

$T_{r+1}.$

Coefficient of these terms are

$^nC_{r-2},\,^nC_{r-1}$ and

$^{n}C_r$ respectively.

These coefficient are equal to

$220,\,495$ and

$792$

$\therefore$

$\dfrac{^nC_{r-2}}{^nC_{r-1}}=\dfrac{220}{495}$

$\Rightarrow$

$\dfrac{r-1}{n-r+2}=\dfrac{4}{9}$

$\Rightarrow$

$9r-9=4n-4r+8$

$\Rightarrow$

$4n+17=13r$ ------ ( 1 )

Also,

$\dfrac{^nC_r}{^nC_{r-1}}=\dfrac{792}{495}$

$\Rightarrow$

$\dfrac{n-r+1}{r}=\dfrac{8}{5}$

$\Rightarrow$

$5n-5r+5=8r$

$\Rightarrow$

$5n+5=13r$

$\Rightarrow$

$5n+5=4n+17$ [ From ( 1 ) ]

$\Rightarrow$

$n=12$

If the last term in the binomial expansion of

$(2{ }^{ \frac { 1 }{ 3 } }-\frac { 1 }{ \sqrt { 2 } } ){ }^{ n }\quad is\quad (\frac { 1 }{ { 3 }^{ \frac { 5 }{ 3 } } } ){ }^{ log{ }_{ 3 }8 }$ , then the 5th term from the beginning is

0%
210
0%
420
0%
105
0%
none of these

Explanation

Given,

$\left ( 2^{\frac{1}{3}}-\dfrac{1}{\sqrt 2} \right )^n$

last term of

$(x-a)^n=(-1)^n(a)^n$

last term of given equation,

$(-1)^n\left ( \dfrac{1}{2^{\frac{n}{2}}} \right )=\left ( \dfrac{1}{3\cdot 9^{\frac{1}{3}}} \right )^{\log_3 8}$

$=\left ( \dfrac{1}{3\cdot 3^{\frac{2}{3}}} \right )^{\log_3 8}$

$=\left ( \dfrac{1}{ 3^{\frac{5}{3}}} \right )^{\log_3 8}$

$=\left ( \dfrac{1}{ 3^{\frac{5}{3}\log_3 8}} \right )$

$=\left ( \dfrac{1}{ 3^{\log_3 8^{\frac{5}{3}}}} \right )$

$=\dfrac{1}{8^{\frac{5}{3}}}$

$=\dfrac{1}{2^5}$

$\Rightarrow \dfrac{n}{2}=5$

$\therefore n=10$

now,

$\left ( 2^{\frac{1}{3}}-\dfrac{1}{\sqrt 2} \right )^{10}$

$T_5=T_{4+1}=^{10}C_4(2^{\frac{1}{3}})^{10-4}\left ( \dfrac{1}{2^{\frac{1}{2}}} \right )^4(-1)^4$

$=^{10}C_4(2^{\frac{6}{3}})\left ( \dfrac{1}{2^{\frac{4}{2}}} \right )$

$=^{10}C_4(2^2)\left ( \dfrac{1}{2^2} \right )$

$=^{10}C_4$

$=210$

The coefficient of

$x^{n}$ in

$(1-x+\frac { { x }^{ 2 } }{ 2! } -\frac { { x }^{ 3 } }{ 3! } +...-\frac { (-1){ }^{ n }x{ }^{ n } }{ n! } ){ }^{ 2 }s\quad equal\quad to\quad$

0%
$\frac { (n){ }^{ n } }{ n! }$
0%
$\frac { (-1)^n(2){ }^{ n } }{ n! }$
0%
$\frac { 1 }{ (n!){ }^{ 2 } }$
0%
$-\frac { 1 }{ (n!){ }^{ 2 } }$

Explanation

Hint: Try using the fact that

$e^{-x}=1-x+\frac{x^{2}}{2!}-\frac{x^{3}}{3!}+\cdots-\frac{(-1)^{n}x^{n}}{n!}$ .

Step 1: Simplication of the given series:

The series expansion of

$e^{-x}$ is given by

$e^{-x}=1-x+\frac{x^{2}}{2!}-\frac{x^{3}}{3!}+\cdots-\frac{(-1)^{n}x^{n}}{n!}$ .

Note that the given expression is the square of

$1-x+\frac{x^{2}}{2!}-\frac{x^{3}}{3!}+\cdots-\frac{(-1)^{n}x^{n}}{n!}$ .

The square of

$1-x+\frac{x^{2}}{2!}-\frac{x^{3}}{3!}+\cdots-\frac{(-1)^{n}x^{n}}{n!}$ would correspond to

$(e^{-x})^{2}$ , that is,

$e^{-2x}$ .

Substitute

$2x$ for

$x$ in

$e^{-x}=1-x+\frac{x^{2}}{2!}-\frac{x^{3}}{3!}+\cdots-\frac{(-1)^{n}x^{n}}{n!}$ to obtain the series expansion of

$e^{-2x}$ .

$\begin{aligned} e^{-2x}&=1-2x+\frac{(2x)^{2}}{2!}-\frac{(2x)^{3}}{3!}+\cdots-\frac{(-1)^{n}(2x)^{n}}{n!}\\e^{-2x}&=1-2x+\frac{4x^{2}}{2!}-\frac{8x^{3}}{3!}+\cdots-\frac{(-1)^{n}(2)^{n}x^{n}}{n!}\end{aligned}$

Step 2 : Calculation of the nth term of the series.

The coefficient of the

$n$ th term in

$e^{-2x}=1-2x+\frac{4x^{2}}{2!}-\frac{8x^{3}}{3!}+\cdots-\frac{(-1)^{n}2^{n}x^{n}}{n!}$ is

$\frac{(-1)^{n}(2)^{n}}{n!}$ .

Final step: The

$n$ th term of the the given series is

$\frac{(-1)^{n}(2)^{n}}{n!}$ .

If in the expansion of

$(1+x)^{20}$ ,the coefficients of rth and (r+4)th terms are equal,then r is equal to

0%
7
0%
8
0%
9
0%
10

Explanation

From given, we have,

$(1+x)^{20}$

and

$r$ th term is equal to

$(r+4)$ th term

$\Rightarrow a=1,b=x,n=20$

$T_{r+1}=^nC_ra^{n-r}b^r$

$\Rightarrow T_{r+1}=^{20}C_r1^{n-r}x^r$

$\therefore T_{r+1}=^{20}C_rx^r$

put

$r=r-1$

$\therefore T_r=^{20}C_{r-1}x^{r-1}$ ..........(1)

put

$r=r+3$

$\therefore T_{r+4}=^{20}C_{r+3}x^{r+3}$ .......(2)

now,

$^{20}C_{r-1}=^{20}C_{r+3}$

we have formula,

$^nC_r=^nC_p\rightarrow r=n-p$

$\Rightarrow r-1=20-(r+3)$

$r-1=20-r-3$

$2r=18$

$\therefore r=9$

${ P }_{ n }$ denotes the product of the binomial coefficients in the expansion of

$(1+x)^{n}$ , then

$\frac { { P }_{ n+1 } }{ { P }_{ n } } =?$

0%
$\cfrac { n+1 }{ n }$
0%
$\cfrac { n^{ n } }{ n }$
0%
$\cfrac {(n+1)^n}{n!}$
0%
none of these

Explanation

$p_n=\,^nC_0.\,^nC_1.\,^nC_2.......^nC_n$

$P_{n+1}=\,^{n+1}C_0.\,^{n+1}C_1.\,^{n+1}C_2.....^{n+1}C_{n+1}$

$\therefore$

$\dfrac{P_{n+1}}{P_n}$

$=\dfrac{\,^nC_0.\,^nC_1.\,^nC_2.......^nC_n}{\,^{n+1}C_0.\,^{n+1}C_1.\,^{n+1}C_2.....^{n+1}C_{n+1}}$

$=\left(\dfrac{^{n+1}C_1}{^nC_0}\right).\left(\dfrac{^{n+1}C_2}{^nC_1}\right).........\left(\dfrac{^{n+1}C_{n+1}}{^nC_n}\right)$

$=\left(\dfrac{\dfrac{n+1}{1}.^nC_0}{^nC_0}\right).\left(\dfrac{\dfrac{n+1}{2}.^nC_1}{^nC_1}\right)......\left(\dfrac{\dfrac{n+1}{n+1}.^nC_n}{^nC_n}\right)$

$=\dfrac{n+1}{1}.\dfrac{n+1}{2}.....\dfrac{n+1}{n+1}$

$=\dfrac{(n+1)^n}{n!}$

The term independent of x in the expansion of

$\displaystyle \left ( x -\dfrac{1}{4} \right )^4\left ( x + \dfrac{1}{x} \right )^3$

0%
19/16
0%
0
0%
14
0%
none of these

$(1+2x+3x^2 )^{10} = a_0+a_1x+a_2x^2+\ ...\,+a_{20}x^{20}$ , then

$a_1$ equals

0%
$10$
0%
$20$
0%
$210$
0%
None of these

Explanation

Given,

$(1+2x+3x^2 )^{10} = a_0+a_1x+a_2x^2+\ ...\ +a_{20}x^{20}$

Then,

$a_1$ = Coefficient of

$x$ in

$(1+2x+3x^2)^{10}$

Now,

$(1+2x+3x^2)^{10}=\{( 1+2x)+3x^2\}^{10}$

$=\ ^{10}C_0 (1+2x)^{10}+\ ^{10}C_1 (1+2x)^9(3x^2)+...$ .

$\therefore$ Coefficient of

$x$ in

$(1+2x+3x^2)^{10}$

$=$ Coefficient of

$x$ in

$^{10}C_0(1+2x)^{10}$

$=\ ^{10}C_0\cdot 2\cdot ^{10}C_1$

$=1\cdot 2\cdot 10=20$

If the coefficients of

$(r-5)^{th}$ and

$(2r-1)^{th}$ terms in the expansion of

$(1+x)^{34}$ are equal, find

$r.$

0%
$21$
0%
$23$
0%
$14$
0%
$20$

Explanation

We know that the coefficient of

$r^{th}$ in the expansion of

${\left(1 + x\right)}^{n}$ is

$^{n}C_{r – 1}$ .

$\therefore\,$ Coefficients of

${\left(r – 5\right)}^{th}$ and

${\left(2r – 1\right)}^{th}$ term in the Expansion of

${\left(1 + x\right)}^{34}$ are

$^{34}C_{r – 6}$ and

$^{34}C_{2r – 2}$

It is given that these coefficients are equal

$^{34}C_{r – 6}=^{34}C_{2r – 2}$

$\Rightarrow\,r – 6 = 2r – 2$ or

$r – 6 + 2r – 2 = 34$

$\left(\because\,^nC_{r}=^{n}C_{S}\Rightarrow\,r = S\,\, or\,\, r + S = n\right)$

$\Rightarrow\,3r – 8 = 34\,\,\,\,\,\,\,\,\left(r – 6 = 2r – 2 \Rightarrow\, r = –4 \,\,which \,\,is\,\, not\,\, possible\right)$

$\Rightarrow\,3r=34+8=42$

$\Rightarrow\,r=\dfrac{42}{3}=14$

Let

$2.^{20}C_0 +^{20} C_1 +^{20}C_2 + ..... +^{20}C_{20}$ . then sum of this series is

0%
$16. 2^{22}$
0%
$8. 2^{20}$
0%
$8. 2^{21}$
0%
$16. 2^{21}$

Explanation

$2. ^{20}C_0 + 5. ^{20} C_1 + 8.^{20} C_2 \, + .... + \, 62. ^{20} C_{20} = \displaystyle \sum_{r = 0}^{20} \, (3r + 2) . ^{20} C_r$

$= 3 \displaystyle \sum_{r = 0}^{20} r. \, ^{20} C_r + 2 \displaystyle \sum_{r = 0}^{20} \, ^{20}C_r = 3 \times 20 \displaystyle \sum_{r = 1}^{20} \, ^{19} C_{r + 1} + 2. (2^{20}) = 60 . 2^{19} + 2.2^{20} = 16.2^{21}$

The greatest term in the expansion of

$(2x + 3y)^{11}$ when x = 9 and y = 4 is :

0%
(165) $(10)^8$ $(12)^3$
0%
(330) $(18)^7 (12)^4$
0%
462 $(18)^8 (12)^5$
0%
None of these

The coefficient of

$x^{18}$ in the product

$(1+x)(1-x)^{10}(1+x+x^2)^9$ is?

0%
$-84$
0%
$84$
0%
$126$
0%
$-126$

Explanation

$\textbf{Step 1 :- Simplify the expression}$

$\text{Given, }\\ \quad (1+x)(1-x)^{10} (1+x+x^2)^9$

$\Rightarrow (1+x)(1-x)^{10}(1+x+x^2)^9 = (1+x)(1-x)(1-x)^9(1+x+x^2)^9$

$\text{We know that, }\\ \quad (a^3-b^3)= (a-b)(a^2+ab+b^2)$

$\Rightarrow (1+x)(1-x)^{10}(1+x+x^2)^9 = (1-x^2)(1-x^3)^9\quad .....(1)$

$\textbf{Step 2 :- Find the coefficient of }\boldsymbol{x^{18}}$

$\text{We know that}\\ \quad \boldsymbol{\textbf{The }{(r+1)}^{th} \textbf{ term in the binomial expansion }(a+b)^n = \ ^nC_r a^{n-r} b^{r}}$

$\text{The }(r+1)^{th} \text{ term in the binomial expansion }(1-x^3)^n = \ ^nC_r \ (-x^3)^{r}= \ ^nC_r (-1)^r x^{3r} \quad ......(2)$

$\text{The coefficient of }x^{18} \text{ in the binomial expansion } (1-x^2)(1-x^3)^9 = \ \text{coefficient of }x^{18} - \text{coefficient of }x^{16} \quad ....(3)$

$\text{The coefficient of }x^{18} \text{ in the binomial expansion } (1-x^3)^9= \ ^9C_6 (-1)^6$

$\text{The coefficient of }x^{16} \text{ in the binomial expansion} (1-x^3)^9 \text{ does not exist. Because there is no r which is integer for}x^{16}.$

$\text{From equation (3) }$

$\text{The coefficient of }x^{18} \text{ in the binomial expansion } (1-x^2)(1-x^3)^9 = \ ^9C_6 (-1)^6 = 84$

$\textbf{Hence, B is the correct option.}$

$p$ is a real number and if the middle term in the expansion of

$\left(\dfrac{p}{2}+2\right)^{8}$ is

$1120$ , find

$p$

0%
$p=\pm 1$
0%
$p=\pm 3$
0%
$p=\pm 5$
0%
$p=\pm 2$

Explanation

Given expansion is

$\left (\dfrac {P}{2}+2\right)^8$

Since index is

$n=8$ , there is only one middle term i.e.,

$\left (\dfrac {8}{2}+1\right)^{th}$

$\Rightarrow \ 5^{th }$ . term

$T_5 =T_{4+1}=^8C_{4} \left(\dfrac {P}{2}\right)^{8-4} 2^4$

$\Rightarrow \ 1120=^{8}C_{4} P^4 \Rightarrow \ 1120=\dfrac {8\times 7\times 6\times 5\times 4!}{4! 4\times 3\times 2\times 1} P^4$

$\Rightarrow \ 1120=7\times 2\times 5\times P^4 =\dfrac {1120}{70}$

$\Rightarrow \ p^4 =16 \ \Rightarrow P^2 =4 \Rightarrow P=\pm 2$

Find the

$7^{th}$ term from the end in the expansion of

$\left(2x^{2}-\dfrac{3}{2x}\right)^{8}$

0%
$4033\ x^{9}$
0%
$4023\ x^{11}$
0%
$4032\ x^{10}$
0%
None of these

Explanation

Given to find the

$7^{th}$ term in expression of

$\left(2x^{2}-\dfrac{3}{2x}\right)^{8}$ from end

$r^{th}$ term from end

$=(n-r+2)^{th}$ term starting

$n=8, r=7$

$7^{th}$ from end

$=(8-7+2)^{th}=3^{rd}$ from starting

$T_{3}=T_{2+1}=\ ^{8}C_{2}(2x^{2})^{6} \left(\dfrac{-3}{2x}\right)^{2}$

$=\dfrac{8!}{2!6!}2^{6}.x^{12}(-1)^{2}\dfrac{3^{2}}{2^{x}x^{2}}$

$T_{3}=4032 x^{10}$

If the middle terms in the expansion of

$\left(x^2+\dfrac{1}{x}\right)^{2n}$ is

$184756x^{10}$ , then what is the value of

$n$ ?

0%
$10$
0%
$8$
0%
$5$
0%
$4$

Explanation

Given,

$\left ( x^2+\dfrac{1}{x} \right )^{2n}$

$2n$ is even

$r=\dfrac{2n}{2} =n$

$MT=^{2n}C_n(x^2)^{2n-n}\left ( \dfrac{1}{x} \right )^n$

$=^{2n}C_n(x^2)^n(x)^{-n}$

$=^{2n}C_n(x)^{2n-n}$

$=^{2n}C_nx^n$

$\therefore 184756x^{10}=^{2n}C_nx^n$

$\Rightarrow x^{10}=x^n$

$\therefore n=10$

Find the coefficient of

$x^{10}$ in the expansion of

$\left(2x^{2}-\dfrac{1}{x}\right)^{20}$

0%
$^{20}C_{8}. 2^{8}$
0%
$^{20}C_{10}. 2^{10}$
0%
$^{20}C_{11}. 2^{11}$
0%
None of these

Explanation

Given that to find the coefficient of

$x^{10}$ in the term

$\left(2x^{2}-\dfrac{1}{x}\right)$

$T_{r+1}=\ ^{20}C_{r}\left.\dfrac{}{}\right| (2x)^{20-r}\left(\dfrac{-1}{x}\right)^{r}$

$=(-1)^{r}\ ^{20}C_{r} 2^{20-r}x^{40-2r-r}$

$=(-1)^{r}\ ^{20}C_{r}2^{20-r}x^{40-3r}$

$40-3r=10\Rightarrow 30=3r\Rightarrow r=10$

$T_{11}=T_{10+1}=(-1)^{10}\ ^{20}C_{10} 20^{20-10}x^{10}$

$=\ ^{20}C_{10} 2^{20-10}x^{10}$

$T_{11}=\ ^{20}C_{10}2^{10}x^{10}$

The coefficient of

$x^4$ in the expansion of

$(1-2x)^5$ is equal to

0%
$40$
0%
$320$
0%
$-320$
0%
$80$

Explanation

General term of

$(1 - 2x)^5$ is given by

$T_{r+1} = \, ^5C_r(-2x)^r$

$= \, ^5C_r(-2)^r x^r$

For coefficient of

$x^4$ , power of

$x = 4$

$\therefore r = 4$

$\therefore$ Coefficient pf

$x^4 = \, ^5C_4(-2)^4$

$= 5\times 16 = 80$

Sum of last

$30$ coefficents in the binomial expansion of

$(1 + x)^{59}$ is

0%
$2^{29}$
0%
$2^{59}$
0%
$2^{58}$
0%
$2^{59} - 2^{29}$
0%
$2^{60}$

Explanation

We have,

$(1 + x)^{59}$

Sum of last

$30$ coefficient of the binomial expansion

$=\ ^{59} C_{30} + \ ^{59} C_{31} + .... + \ ^{59}C_{59}$

We know that,

$^{59} C_0 + \ ^{59}C_1 + \ ^{59}C_2 + .... + \ ^{59} C_{59} = 2^{59}$

$\Rightarrow (^{59}C_0 +\ ^{59}C_{59}) + (^{59} C_1 +\ ^{59}C_{58}) + ..... + (^{59} C_{29} + \ ^{59}C_{30}) = 2^{59}$

$\Rightarrow 2 (^{59} C_{59} + \ ^{59}C_{58} + .... + \ ^{59}C_{31} + \ ^{59}C_{30}) = 2^{59}$

$[\because \ ^nC_r = \ ^nC_{n - r}]$

$\Rightarrow \ ^{59} C_{30} +\ ^{59} C_{31} + .... \ ^{59} C_{39} = \dfrac{2^{59}}{2} = 2^{58}$

$\therefore$ Sum of last

$30$ coefficient of the binomial expansion

$(1 + x)^{59}$ is

$2^{58}$ .

Find the coefficient of

$x^{-15}$ in the expansion of

$\left(3x^{2}-\dfrac{a}{3x^{3}}\right)^{10}$

0%
$-\dfrac {40}{21}a^6$
0%
$-\dfrac {30}{23}a^8$
0%
$-\dfrac {40}{27}a^7$
0%
None of these

Explanation

Given that, to find coefficient of

$x^{-15}$ in equation of

$\left (3x^2 -\dfrac {a}{3x^2}\right)^{10}$

$T_{r+1}=^{10}C_{r} (3x)^{10-r} \left (\dfrac {-a}{3x^3}\right)^{10}$

$=(-1) ^{10}C_{r}(3)^{10-r} (x^{20-2r}) (a^r) (x^{-3r}) \left (\dfrac {1}{3}\right)^r$

$=(-1)^r\ ^{10}C_{r} 3^{10-r-r} x^{20-2r-3r} a^r$

$T_{r+1}=(-1)^r\ ^{10}C_{r} 3^{10-2r} x^{20-5r} a^r$

$20-5r =-15\ \Rightarrow 35=5r \Rightarrow r=7$

$T_{8}=T_{7+1}=(-1)^7\ ^{10}C_{7} 3^{10-2(7)} x^{20-5(7)} a^7$

$=-^{10}C_{7} 3^{r} x^{-15} a^7 \Rightarrow -\dfrac {40}{27}a^7 x^{-15}$

Coefficient of

$x^{-15}$ is

$-\dfrac {40}{27}a^7$

If the sum of the coefficients in the expansions of

$(a^2 x^2 - 2ax + 1)^{51}$ is zero, then

$a$ is equal to

0%
$0$
0%
$1$
0%
$-1$
0%
$-2$
0%
$2$

Explanation

$(a^2 x^2 - 2ax + 1)^{51}$

For sum of coefficients put

$x = 1$

$(a^2 - 2a + 1)^{51} = 0$

$\Rightarrow \{(a - 1)^2 \}^{51} = 0$

$\Rightarrow a = 1$

$\dfrac{1}{9!} + \dfrac{1}{3! 7!} + \dfrac{1}{5! 5!} + \dfrac{1}{7! 3!} + \dfrac{1}{9!}$ is equal to

0%
$\dfrac{2^9}{10!}$
0%
$\dfrac{2^{10}}{8!}$
0%
$\dfrac{2^{11}}{9!}$
0%
$\dfrac{2^{10}}{7!}$
0%
$\dfrac{2^{8}}{9!}$

Explanation

$\dfrac{1}{9!} + \dfrac{1}{3! 7!} + \dfrac{1}{5! 7!} + \dfrac{1}{7! 3!} + \dfrac{1}{9!}$

$=\dfrac{1}{10!}\left[\dfrac{10}{9!1!} + \dfrac{10}{3! 7!} + \dfrac{10}{5! 7!} + \dfrac{10}{7! 3!} + \dfrac{10}{9!1!}\right]$

$=\dfrac{1}{10!}[^{10}C_1+\,^{10}C_3+\,^{10}C_5+\,^{10}C_7+\,^{10}C_9]$

$=\dfrac{1}{10!}.2^9=\dfrac{2^9}{10!}$

The arithmetic mean of

$^nC_0 , \ ^nC_1, \ ^nC_2 ...., \ ^nC_n$ is

0%
$\dfrac{2^n}{n + 1}$
0%
$\dfrac{2^n}{n}$
0%
$\dfrac{2^{n -1}}{n + 1}$
0%
$\dfrac{2^{n - 1}}{n}$
0%
$\dfrac{2^{n + 1}}{n}$

Explanation

$\because (1 + x)^n = \ ^nC_0 + \ ^nC_1 x + \ ^nC_2 . x^2 + ... + \ ^nC_n . x^n$

Take

$x = 1$

$(1 + 1)^n = \ ^nC_0 + \ ^nC_1 . (1) + \ ^nC_2 . (1)^2 + ... + \ ^nC_0 . (1)^n$

$2^n = \ ^nC_0 + \ ^nC_1 + \ ^n C_2 + ... + \ ^nC_n$

Now arithmetic mean

$\bar{X} = \dfrac{\displaystyle \sum_{i = 1}^n x_i}{N}$ , where

$N = (n + 1)$

$\Rightarrow \bar{X} = \dfrac{^nC_0 + \ ^nC_1 + \ ^n C_2 + .... + \ ^nC_n}{n + 1}$

$\Rightarrow \bar{X} = \dfrac{2^n}{n+ 1}$

The largest term in the expansion of

$(2 + 3x)^{25}$ where x = 2 is its

0%
13th term
0%
19th term
0%
20th term
0%
26th term

In the expansion of

$(1 + x)^{43}$ , the coefficients of the (2r+1)th and the (r + 2)th terms are equal, then the value of r, is

0%
14
0%
15
0%
16
0%
17

The total number of terms in the expansion of

$(x + a)^{100} + (x - a)^{100}$ after simplification,

0%
50
0%
51
0%
154
0%
202

The largest term in the expansion of

$\left(\frac{b}{2} + \frac{b}{2}\right)^{100}$ is

0%
$b^{100}$
0%
$\left(\frac{b}{2}\right)^{100}$
0%
$^{100}C_{50} \left(\frac{b}{2}\right)^{100}$
0%
None of these

Which of the following expansion will have term containing

$x^2$

0%
$\displaystyle(x^{-1/5} + 2x^{3/5})^{25}$
0%
$\displaystyle(x^{3/5} + 2x^{-1/5})^{24}$
0%
$\displaystyle (x^{3/5} - 2x^{-1/5})^{23}$
0%
$\displaystyle (x^{3/5} + 2x^{-1/5})^{22}$

$\displaystyle ( 1+ 2x + 3x^2)^{10} = a_0 + a_1x + a_2x^2 + \dots + a_{20}x^{20}$ then

$a_1$ equals

0%
10
0%
20
0%
210
0%
420

The coefficient if

$x^6$ in the expansion of

$( 3x^2 - \frac {1}{3x})^9$ is

0%
378
0%
756
0%
189
0%
567

If the second term in the expansion

$\displaystyle \lgroup ^{13}\sqrt{a} + \frac{a}{\sqrt{a^{-1}}} \rgroup^n$ is

$\displaystyle 14a^{5/2}$ , then the value of

$\displaystyle ^nC_3/^nC_2$ is

0%
4
0%
3
0%
12
0%
6

If the fourth term of

$\displaystyle \left(\sqrt{_x\left(\frac{1}{1 + \log_{10} x}\right)} + ^{12}\sqrt{x} \right)^6$ is equal to 200 and x > 1, then x is equal to

0%
$\displaystyle 10\sqrt{2}$
0%
$\displaystyle 10$
0%
$\displaystyle 10^4$
0%
None of these

The sum of rational term in

$(\sqrt 2+\sqrt [3]{3}+\sqrt[6]{5})^{10}$ is equal to

0%
$12632$
0%
$1260$
0%
$1236$
0%
none of these

Explanation

General term in the expansion of

$( \sqrt 2+ \sqrt[3]{3}+\sqrt[6]{5})^{10}$ is

$\dfrac{10!}{a!b!c!}(\sqrt 2)^a (\sqrt[3]{3})^b ( \sqrt[6]{5})^c$ where

$a+b+c=10$ .
For rational term, we have the following:

Value of $a, b, c$	Value of term
$a=4, b=0, c=6$	$\dfrac{10!}{4!0!6!}( \sqrt 2)^4 (\sqrt[3]{3})^0( \sqrt [6]{5})^6=4200$
$a=10, b=0, c=0$	$\dfrac{10!}{10!0!0!}(\sqrt 2)^{10}( \sqrt[3]{3})^0 (\sqrt [6]{5})^0=32$
$a=4, b=6, c=0$	$\dfrac{10!}{4!6!0!}(\sqrt 2)^{4}( \sqrt[3]{3})^6 (\sqrt [6]{5})^0=7560$

The number of distinct terms in the expansion of

$\left( x+\dfrac{1}{x}+x^2+\dfrac{1}{x^2}\right)^{15}$ is/are ( with respect to different power of

$x$ )

0%
$255$
0%
$61$
0%
$127$
0%
none of these

The general term in the expansion of

$(x + a)^n$

0%
$\,^nC_r\,x^{n-r} .a^r$
0%
$\,^nC_r\,x^{r} .a^r$
0%
$\,^nC_{n-r}\,x^{n-r} .a^r$
0%
$\,^nC_{n-r}\,x^{r} .a^{n-r}$

Explanation

The general term in the expansion of

$(x + a)^n$ is

$\,^nC_r\,x^{n-r} .a^r$

The

$7th$ term in the expansion of

$\bigg(\dfrac{1}{2}+a\bigg)^8$ is :

0%
$\,^8C_7\,\bigg(\dfrac{1}{2}\bigg)(a)^7$
0%
$\,^8C_7\,\bigg(\dfrac{1}{2}\bigg)^7.a$
0%
$\,^8C_6\,\bigg(\dfrac{1}{2}\bigg)^2(a)^6$
0%
$\,^8C_6\,\bigg(\dfrac{1}{2}\bigg)^6(a)^2$

Explanation

$7th$ term in the expansion of

$\bigg(\dfrac{1}{2}+a\bigg)^8$

$=T_7=T_6+1$

$=\,^8C_6\,\bigg(\dfrac{1}{2}\bigg)^{8-6}(a)^6$

$(\because T_{r+1}=\,^nC_r\,a^{n-r}\,b^r)$

$=\,^8C_6\,\bigg(\dfrac{1}{2}\bigg)^2(a)^6$

Find the middle term of the expansion of

$\left ( 3x+\frac{1}{2x} \right )^7$

0%
$^{7}C_{4}\dfrac{2^5 }{16 x^2}$
0%
$^{7}C_{3}\dfrac{2^7 }{16 x}$
0%
$^{7}C_{4}\dfrac{2^7 }{16 x}$
0%
$^{-7}C_{3}\dfrac{2^7 }{16 x}$

The sum of the coefficients of even powers of

$x$ in the expansion of

$(1+x+x^2+x^3)^5$ is

0%
$512$
0%
$256$
0%
$128$
0%
$64$

Explanation

Let

$f(x)=(1+x+x^2+x^3)\:^5$

$=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3...$ ...(i)

$f(1)=a_{0}+a_{1}+a_{2}...$ ...(ii)

$f(-1)=a_{0}-a_{1}+a_{2}-a_{3}...$ ...(iii)
Adding (ii) and (iii), we get

$f(1)+f(-1)=2[a_{0}+a_{2}+a_{4}...]$

$(1+1+1+1)^{5}+(1-1+1-1)^{5}=2[a_{0}+a_{2}+a_{4}...]$

$4^{5}=2[a_{0}+a_{2}+a_{4}...]$

$2^{10}=2[a_{0}+a_{2}+a_{4}...]$

$2^9=a_{0}+a_{2}+a_{4}...$

$512=a_{0}+a_{2}+a_{4}...$
Hence sum of the coefficients of even powers of

$x$ is

$512$ .

The coefficient of

${x}^{5}$ in the expansion of

$(1+x)^{21}+(1+x)^{22}+\ldots\ldots..+(1+x)^{30}$ is

0%
$^{51}{{C}_{5}}$
0%
$^{9}{{C}_{5}}$
0%
$^{31}{{C}_{6}-}^{21}{{C}_{6}}$
0%
$^{30}{{C}_{5}+}^{20}{{C}_{5}}$

Explanation

Let

$S=(1+x)^{21}+(1+x)^{22}+(1+x)^{23}+...(1+x)^{30}$ ...(i)

$(1+x)S=(1+x)^{22}+(1+x)^{23}...(1+x)^{30}+(1+x)^{31}$ ...(ii)

Subtracting

$(i)$ from

$(ii)$ , we get

$xS=(1+x)^{31}-(1+x)^{21}$

$S=\dfrac{(1+x)^{31}}{x}-\dfrac{(1+x)^{21}}{x}$

Coefficient of

$x^{r}$

$=\:^{31}C_{r}x^{r-1}-\:^{21}C_{r}x^{r-1}$

$=(\:^{31}C_{r}-\:^{21}C_{r})x^{r-1}$ ...

$(a)$

Therefore the coefficient of

$x^5$ implies

$r-1=5$

$r=6$

Substituting in

$(a)$ , we get coefficient of

$x^5$

$=(\:^{31}C_{6}-\:^{21}C_{6})$

The coefficient of middle term in the expansion of

$(1+{x})^{40}$ is

0%
$\displaystyle \frac{1.3.5\cdots 39}{20!}.2^{20}$
0%
$\displaystyle \frac{1.3.5\cdots 39}{20!}$
0%
${\dfrac{40}{20}!}$
0%
$40!$ $20^{20}$

Explanation

${\textbf{Step - 1: Determine the Middle term}}$

${\text{Middle term = }}\dfrac{{\text{N}}}{2} + 1 = \dfrac{{2{\text{n}}}}{2} + 1$

$= \left( {{\text{n + 1}}} \right){\text{th term}}$

${\text{The middle term of }}{\left( {{\text{1 + x}}} \right)^{40}}{\text{ is }}\left( {\dfrac{{40}}{2} + 1} \right){\text{ i}}{\text{.e 2}}{{\text{1}}^{th}}{\text{ term in that expansion}}$

${{\text{T}}_{21}} = {{\text{ }}^{40}}{{\text{C}}_{20}}{{\text{x}}^{40 - 20}} = {{\text{ }}^{40}}{{\text{C}}_{20}}{{\text{x}}^{20}}$

${\text{Therefore the coefficient of the middle term is}}{{\text{ }}^{40}}{{\text{C}}_{20}}$

$\therefore {{\text{ }}^{40}}{{\text{C}}_{20}} = \dfrac{{40!}}{{20!\left( {40 - 20} \right)!}} = \dfrac{{40!}}{{20!20!}}$

${\textbf{Step - 2 : Simplify the term}}{\text{.}}$

${\text{Dividing by 20!;and separating the multiples of two in the powers of two we get,}}$

$\therefore {\text{Middle term = }}\dfrac{{\left( {1.3.5.7.....39} \right){2^{20}}}}{{20!}}$

${\textbf{Hence, Correct answer will be }}\dfrac{{\left( {1.3.5.7.....39} \right){2^{20}}}}{{20!}}$

$S$ be the sum of the coefficients in the expansion of

$(ax+by+-cz)^{n}$ where

$a, b, c$ are lengths of the sides of a triangle, then

$\lim_{n\rightarrow \infty }\dfrac{S}{(S^{1/n}+1)^{n}}$ is

0%
$1$
0%
$0$
0%
$e^{\left(\dfrac{ab}{c}\right)}$
0%
$e^{\displaystyle \left(\frac{a+b+c}{a+b+c-1}\right)}$

Explanation

$S=(a+b-c)^{n}$

$\displaystyle \lim_{n\rightarrow \infty }\dfrac{S}{\left ( S^{1/n}+1 \right )^{n}}$ =

$\displaystyle \lim_{n\rightarrow \infty }\left ( \dfrac{a+b-c}{a+b-c+1} \right )^{n}$

$\displaystyle =\lim_{n\rightarrow \infty } \left ( \dfrac{k}{k+1} \right )^{n} = L$ (say)
Where

$k = a + b - c$
Now In

$\Delta ABC a + b > c$

$\Rightarrow 0< \dfrac{k}{k+1}< 1 \therefore L = 0$

Match the elements of List I with List II

List I	List II
A) lf $\lambda$ be the number of terms which are integers, in the expansion of $(5^{\frac16}+7^{\frac19})^{1824}$ , then $\lambda$ is divisible by	P) 2
B) lf $\lambda$ be the number of terms which are rational in the expansion of $(5^{\frac16}+2^{\frac18})^{100}$ , then $\lambda$ is divisible by	Q) 3
C) lf $\lambda$ be the number of terms which are irrational in the expansion of $(3^{\frac14}+4^{\frac13})^{99}$ , then $\lambda$ is divisible by	R) 7
	S) 13
	T) 17

The correct option which matches all the elements correctly, is :

0%
(A) - P,Q,T (B) - P (C) - R,S
0%
(A) - P,T (B) - Q (C) - S,T
0%
(A) - P,S,T (B) - S,Q (C) - Q
0%
(A) - P,S (B) - P,A (C) - P,R

Explanation

(i) $(5^{\frac{1}{6}}+7^{\frac{1}{9}})^{1824}$

$T_{r+1}=^{1824}C_r (5)^{\frac{1824-r}{6}} 7^{\frac{r}{9}}$

For integer terms, $r$ should be multiple of $9$ .

For $r=18,36,54,72,.....1818$ , terms comes as integer.

This is an A.P.

$1818=18+(n-1)18$

$\Rightarrow n=101$

Also, for $r=0$ , we would get an integer

So, total number of terms which gives integer values are $101+1=102.$

So, $\lambda=102$

So, $\lambda$ is divisible by $2,3,17$

(ii) $(5^{\frac{1}{6}}+2^{\frac{1}{8}})^{1824}$

$T_{r+1}=^{100}C_r (5)^{\frac{100-r}{6}}2^{\frac{r}{8}}$

For rational terms, $r$ should be multiple of $8$ .

For $r=16,40,64,88$ , terms comes as rational.

So, number of rational terms are $4$ .

So, $\lambda=4$

which is divisble by $2$ .

(iii) $(3^{\frac{1}{4}}+4^{\frac{1}{3}})^{99}$

$T_{r+1}=^{99}C_r (3)^{\frac{99-r}{64}}4^{\frac{r}{3}}$

For rational terms, $r$ should be multiple of $3$ .

For $r=3,15,27,.....97$ , terms comes as rational.

This is an AP

$97=3+(n-1)12$

$\Rightarrow n=8$

For $r=99$ also, there is a rational value

So, number of rational terms are $8+1=9$

Now, number of irrational terms = total number of terms -rational number of terms

$=99+1-9=91$

So, $\lambda=91$

which is divisble by $7,13$ .

Hence, option A is the correct answer.

Arrange the values of

$n$ in ascending order
A : If the term independent of

$x$ in the expansion of

$\left(\displaystyle \sqrt{x}-\frac{n}{x^{2}}\right)^{10}$ is

$405$
B : If the fourth term in the expansion of

$\left(\displaystyle \frac{1}{n}+n^{\log_{n}10}\right)^{5}$ is

$1000$ , (

$n< 10$ )
C : In the binomial expansion of

$(1+x)^{n}$ the coefficients of

$5^{\mathrm{t}\mathrm{h}},\ 6^{\mathrm{t}\mathrm{h}}$ and

$7^{\mathrm{t}\mathrm{h}}$ terms are in A.P.

0%
$A,B,C$
0%
$B,A,C$
0%
$A,C,B$
0%
$C,A,B$

Explanation

A: Given:

$(\sqrt{x}-nx^{-2})^{10}$

$T_{r+1}=(-1)^{r}\:^{10} C_{r} x^{(5-{5r}/{2})} n^{r}$ ...(1)

Term independent of

$x$ implies

$5-\dfrac{5r}{2}=0$

Thus

$r=2$

Substituting in equation (1), we get

$T_{3}=\:^{10}C_{2}n^{2}=405$

$\Rightarrow 45n^{2}=405$

$\Rightarrow n^{2}=9$

$\Rightarrow n=3$ (

$n$ is

$\neq -3$ as

$n$ is a positive integer)

B: Given:

$(n^{-1}+n^{log_{n}10})^{5}$

$=(n^{-1}+10)^{5}$

$T_{r+1}=\:^{5}C_{r}n^{5-r}10^{r}$ ...(2)
It is given that the third term is

$1000$
Substituting in equation (2), we get

$T_{4}=\:^{5}C_{2}n^{-2}10^{3}=1000$

$\Rightarrow 10(n^{-2}1000)=1000$

$\Rightarrow n^{-2}=10$

$\Rightarrow n=\dfrac{1}{\sqrt{10}}$

C: For the terms to be in arithmetic progression

$(n-2r)^{2}=n+2$ ...(3)
Let us consider

$\:^{n}C_{r-1},\:^{n}C_{r},\:^{n}C_{r+1}$ be in A.P.
Hence,

$T_{r},T_{r+1},T_{r+2}$ are in A.P.
Here it is given that, the

$5^{th}, 6^{th}$ and

$7^{th}$ terms are in A.P.
Hence,

$r=5$
Substituting in equation (3), we get

$(n-10)^{2}=n+2$

$\Rightarrow n^{2}-21n+98=0$

$\Rightarrow(n-7)(n-14)=0$
Hence,

$n=7, 14$
Hence, the values of

$n$ in ascending order is B,A,C.

Option B.

The value of the expression

$\displaystyle \frac{C_1}{C_0}+2\frac{C_2}{C_1}+3\frac{C_3}{C_2}+\ldots\ldots\ldots +n\frac{C_n}{C_{n-1}}$ is

0%
$\dfrac{(n+1)(n+2)}{2}$
0%
$\dfrac{n(n+1)}{2}$
0%
$\dfrac{n(n-1)}{2}$
0%
$\dfrac{n(n+2)}{2}$

Explanation

Given:

$\dfrac{C_1}{C_0} + 2\dfrac{C_2}{C_1}+3\dfrac{C_3}{C_2} +\cdots+n\dfrac{C_n}{C_{n-1}}$

$r^{th}$ term of the above expression is

$T_r = r\dfrac{C_r}{C_{r-1}} = r\dfrac{n-r+1}{r} = n-r+1$

Now,

$\displaystyle\sum_{r=1}^nT_r = \sum_{r=1}^n (n-r+1)$

$=n^2-\dfrac{n(n+1)}{2} + n$

$=\dfrac{n(n+1)}2$

The coefficient of

$x^r[0 \le r \le n-1]$ in the expression of

$(x + 2)^{n-1} + (x+2)^{n-2} .(x+1) + (x+2)^{n-3}. (x+1)^2+...+(x+1)^{n-1}$ is

0%
$^nC_r(2^r - 1)$
0%
$^nC_r(2^{n-r} - 1)$
0%
$^nC_r(2^r + 1)$
0%
$^nC_r(2^{n-r} + 1)$

Explanation

Given expression is equal to

$\sum _{ p=1 }^{ n }{ { (x+3) }^{ (n-p) }\times { (x+2) }^{ (p-1) } }$

Which is equal to

${ (x+3) }^{ (n-1) }\times (\sum _{ p=1 }^{ n }{ { (\dfrac { (x+2) }{ (x+3) } ) }^{ (p-1) }) } = { (x+3) }^{ n }-{ (x+2) }^{ n }$

So,coefficient of

${ x }^{ r } \space is \space^n{ C }_{ r }\times ({ 3 }^{ (n-r) }-{ 2 }^{ (n-r) })$

$\cfrac { { C }_{ 0 } }{ x } -\cfrac { { C }_{ 1 } }{ x+1 } +\cfrac { { C }_{ 2 } }{ x+2 } -......+{ \left( -1 \right) }^{ n }\cfrac { { C }_{ n } }{ x+n } =$ _______ where

${ C }_{ r }$ stands for

${ _{ }^{ n }{ C } }_{ r }$ .

0%
$\cfrac { n! }{ (x+1)...(x+n) }$
0%
$\cfrac { n! }{ x(x+1)...(x+n-1) }$
0%
$\cfrac { n! }{ x(x+1)...(x+n) }$
0%
$\cfrac { n-1! }{ x(x+1)...(x+n) }$

Explanation

$\dfrac { { c }_{ 0 } }{ x } -\dfrac { { c }_{ 1 } }{ x+1 } +\dfrac { { c }_{ 2 } }{ x+2 } -............\quad \quad +{ \left( -1 \right) }^{ n }\dfrac { { c }_{ n } }{ x+n }$

${ \left( 1-y \right) }^{ n }={ c }_{ 0 }{ - }{ c }_{ 1 }y+{ c }_{ 2 }{ y }^{ 2 }- ......+{ c }_{ n }{ y }^{ n }$

${ y }^{ x-1 }{ \left( 1-y \right) }^{ n }={ c }_{ 0 }{ y }^{ x-1 }-{ c }_{ 1 }{ y }^{ x }+{ c }_{ 2 }{ y }^{ x+1 }-......$

$\displaystyle \int _{ 0 }^{ 1 }{ { y }^{ x-1 } } { \left( 1-y \right) }^{ n }dy=\int _{ 0 }^{ 1 }{ { c }_{ 0 }{ y }^{ x-1 } } dy-\int _{ 0 }^{ 1 }{ { c }_{ 1 }{ y }^{ x }dy } +..........$

$\Rightarrow \displaystyle \int _{ 0 }^{ 1 }{ { y }^{ x-1 } } { \left( 1-y \right) }^{ n }dy=\dfrac { { c }_{ 0 } }{ x } -\dfrac { { c }_{ 1 } }{ x+1 } +.......+{ \left( -1 \right) }^{ n }\dfrac { { c }_{ n } }{ x+n }$

We have to calculate the the value of integration in order to find the required value of RHS.

$\displaystyle \int _{ 0 }^{ 1 }{ \dfrac { { y }^{ x-1 } }{ 1 } } { \left( 1-y \right) }^{ n } dy$ ...... (using by parts)

$=\displaystyle { \left[ \dfrac { { y }^{ x } }{ x } { \left( 1-y \right) }^{ n } \right] }_{ 0 }+n\int _{ 0 }^{ 1 }{ \dfrac { { y }^{ x } }{ x } { \left( 1-y \right) }^{ n-1 } } dy$

$=\displaystyle 0+\dfrac { n }{ x } \int _{ 0 }^{ 1 }{{ y }^{ x } { \left( 1-y \right) }^{ n-1 } } dy$

Again by using by parts:-

$=\displaystyle \dfrac { n }{ x } \cdot { \left[ \dfrac { { y }^{ x+1 } }{ x+1 } \cdot { \left( 1-y \right) }^{ n-1 } \right] }_{ 0 }+\int _{ 0 }^{ 1 }{ \left( n-1 \right) { \left( 1-y \right) }^{ n-2 }\dfrac { { y }^{ x+1 } }{ x+1 } }$

$=\displaystyle \dfrac { n }{ x } \cdot \left[ 0+\dfrac { n-1 }{ x+1 } \int _{ 0 }^{ 1 }{ { y }^{ x+1 }{ \left( 1-y \right) }^{ n-2 } } dy \right]$

$=\displaystyle \dfrac { n\left( n-1 \right) }{ x\left( x+1 \right) } \int _{ 0 }^{ 1 }{ { y }^{ x+1 } } { \left( 1-y \right) }^{ n-2 }dy$

This integration by parts will go untill

${ \left( 1-y \right) }^{ n-2 }$ becomes

${ \left( 1-y \right) }^{ 0 }$ i.e. 1 and then the integration will stop.

So, final integration will be

$\displaystyle \int _{ 0 }^{ 1 }{ { y }^{ x+n-1 } } dy = \dfrac { { y }^{ x+n } }{ x+n } =\dfrac { 1 }{ x+n }$

Thus,

$\displaystyle \int_{0}^{1} y^{x-1} (1-y)^{n} = \dfrac { n\left( n-1 \right) \left( n-2 \right) .....1 }{ x\left( x+1 \right) \left( x+2 \right) ....\left( x+n \right) } =\dfrac { n! }{ x\left( x+1 \right) ...\left( x+n \right) }$

Hence, the correct option is C.

${ \left( 1+x \right) }^{ n }={ C }_{ 0 }+{ C }_{ 1 }{x}+{ C }_{ 2 }{ x }^{ 2 }+\cdot \cdot \cdot \cdot \cdot +{ C }_{ n }{ x }^{ n }$ , then

0%
$\cfrac { { C }_{ 1 } }{ { C }_{ 0 } } +2\cfrac { { C }_{ 2 } }{ { C }_{ 1 } } +3\cfrac { { C }_{ 3 } }{ { C }_{ 2 } } +\cdot \cdot \cdot \cdot +n\cfrac { { C }_{ n } }{ { C }_{ n-1 } } =\cfrac { n(n+1) }{ 2 }$
0%
${ C }_{ n-1 } = n$
0%
$\cfrac { { C }_{ 1 } }{ { C }_{ 0 } } +2\cfrac { { C }_{ 2 } }{ { C }_{ 1 } } +3\cfrac { { C }_{ 3 } }{ { C }_{ 2 } } +\cdot \cdot \cdot \cdot +n\cfrac { { C }_{ n } }{ { C }_{ n-1 } } =\cfrac { n(n-1) }{ 2 }$
0%
None of the above

Explanation

$\displaystyle \frac{^{n}\textrm{C}_{1}}{^{n}\textrm{C}_{0}}+2\displaystyle \frac{^{n}\textrm{C}_{2}}{^{n}\textrm{C}_{1}}+3\displaystyle \frac{^{n}\textrm{C}_{3}}{^{n}\textrm{C}_{2}}+...$

$=\displaystyle \frac{n}{1}+\displaystyle \frac{2(n-1)}{2}+\displaystyle \frac{3(n-2)}{3}+...$

$=n+(n-1)+(n-2)+... +1$

$=n.n-(1+2+....+(n-1))$

$=\displaystyle\frac{n(n+1)}{2}$

Also,

$^{n}\textrm{C}_{n-1}=\displaystyle \frac{n!}{(n-1)!1!}=n$

Hence, options A and B are the correct answers.

If the

$(n+1)$ numbers

$a,b,c,d,...$ be all different and each of them a prime number, then the number of different factors (other than 1) of

$a^m.b.c.d....$ is

0%
$m-2^n$
0%
$(m+1)2^n$
0%
$(m+1)2^n-1$
0%
None of these

Explanation

No. of elements in

$b.c.d... = n$
Choose

$a^k$ , where k

$\in {0,1,2, ..m}$ at a time =

$(m+1)$
for every

$k$ , no. possible factors from n elements =

$^nC_0+^nC_1+^nC_2+ . . . . . + ^nC_n$

$= 2^n$
Total factors

$=$ No. of possible powers of

$a\times$ every

$k$ , no. possible factors from

$n$ elements

$= (m+1)2^n$
If factor

$1$ is excluded then

$(m+1)2^n-1$

Find the sum of the series

$3.{ _{ }^{ n }{ C } }_{ 0 }-8.{ _{ }^{ n }{ C } }_{ 1 }+13._{ }^{ n }{ { C }_{ 2 } }-18.{ _{ }^{ n }{ { C }_{ 3 } } }+\ldots+(n+1)\quad terms$

0%
$0$
0%
$-1$
0%
$+1$
0%
None of these

Explanation

Let

$n=2$
Hence the above expression is reduced to

$3(1)-8(2)+13(1)$

$=16-16=0$
Let

$n=3$

$3(1)-8(3)+13(3)-18(1)$

$=42-42=0$
Hence the sum of the series for

$n>1$ is

$0$

if the coefficient of the middle term in the expansion of

$\displaystyle (1+x)^{2n+2}$ and

$p$ and the coefficients of middle terms in the expansion of

$\left ( 1+x \right )^{2n+1}$ are

$q$ and

$r$ ,then

0%
$\displaystyle p+q=r$
0%
$\displaystyle p+r=q$
0%
$\displaystyle p=q+r$
0%
$\displaystyle p+q+r=0$

Explanation

$(1+x)^{2n+2}$ .
The middle term will be $(\dfrac {N}{2}+1)$ th term.
$=\dfrac {2n+2}{2}+1$ th term
$=(n+2)^{th}$ term
Hence coefficient of the middle term will be
$\:^{2n+2}C_{n+1}$
$=p$
For
$(1+x)^{2n+1}$ .
The middle terms will be $(\dfrac {N+1}{2}+1)$ and $(\dfrac {N+1}{2})$ terms
$=\dfrac {2n+2}{2}+1$ th term and $\dfrac {2n+2}{2}^{th}$ term.
$=(n+2)^{th}$ term and $(n+1)^{th}$ term.
Hence coefficient of the middle terms will be
$\:^{2n+1}C_{n+1}=q$ and $\:^{2n+1}C_{n}=r$
By the properties of binomial coefficients.
$\:^{2n+1}C_{n+1}+\:^{2n+1}C_{n}$
$=\:^{2n+2}C_{n+1}$
Hence $q+r=p$

In the expansion of

$(5^{\tfrac 12} + 2^{\tfrac 18})^{1024}$ , the number of integral terms is

0%
$128$
0%
$129$
0%
$130$
0%
$131$

Explanation

$\displaystyle T_{r+1} = ^{1024}C_r\times 5^{\tfrac r2 }\times2^{\tfrac {(1024-r)}8}$
For

$T_{r+1}$ to be integral, the powers of

$5$ and

$2$ must be integers.
Hence,

$\displaystyle\frac {r}{2} = a$ and

$\displaystyle\frac{1024-r}{8} = b$ , where

$a$ and

$b$ are integers.
Hence,

$r$ must be a factor of

$8$ .

$r$ can assume values from

$0$ to

$1024$ which are factors of

$8$ .
Hence, the total number of values

$r$ can acquire is

$1 + \displaystyle\frac{1024}{8} = 129$ .

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 10 - MCQExams.com

0:0:2

Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 10 - MCQExams.com

0:0:2

Practice Class 11 Engineering Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.