MCQ Questions for Class 11 Engineering Maths Binomial Theorem Quiz 1

The sum of coefficient of integral powers of

$x$ in the binomial expansion of

$(1-2\sqrt x)^{50}$ is :

0%
$\dfrac {1}{2}(3^{50}+1)$
0%
$\dfrac {1}{2}(3^{50})$
0%
$\dfrac {1}{2}(3^{50}-1)$
0%
$\dfrac {1}{2}(2^{50}+1)$

Explanation

$(1-2\sqrt{x})^{50}= ^{50}C_0-^{50}C_1(2\sqrt{x})+^{50}C_2(2\sqrt{x})^2.....$

we need to find sum of even forms, so

Put

$\sqrt{x}=1 \Longrightarrow (1-2)^{50}=1=^{50}C_0-^{50}C_1(2)+^{50}C_2(2)^2.........(1)$

and

$\sqrt{x}=-1 \Longrightarrow (1+2)^{50}=3^{50}=^{50}C_0+^{50}C_12+^{50}C_2(2)^2.......(2)$

Adding (1) and (2) we get,

$3^{50}+1=2(^{50}C_0+^{50}C_2(2)^2+........)$

$\Rightarrow (^{50}C_0+^{50}C_2(2)^2+........)=\cfrac{1}{2}(1+3^{50})$

$(2+\dfrac {x}{3})^{55}$ is expanded in the ascending powers of

$x$ and the coefficients of powers of

$x$ in two consecutive terms of the expansion are equal, then these terms are :

0%
$7^{th}$ and $8^{th}$
0%
$8^{th}$ and $9^{th}$
0%
$28^{th}$ and $29^{th}$
0%
$27^{th}$ and $28^{th}$

Explanation

Coefficient of $x^r$ and $x^{r+1}$ are equal.

Coefficient of $T_{r+1} =$ Coefficient of $T_{r+2}$

$\Rightarrow\displaystyle { ^{ 55 }C }_{ r }{ (2) }^{ 55-r }{ \left(\dfrac { 1 }{ 3 } \right) }^{ r }={ ^{ 55 }C }_{ r+1 }{ (2) }^{ 54-r }{ \left(\dfrac { 1 }{ 3 } \right) }^{ r+1 }$

.
$\dfrac { 55! }{ (55-r)!r! } (2)=\dfrac { 55! }{ (54-r)!(r+1)! } \dfrac { 1 }{ 3 }$

$\Rightarrow 6(r+1) = 55-r$

$\Rightarrow r =7$

So, the terms will be $8^{th}$ and $9^{th}$

If the sum of odd numbered terms and the sum of even numbered terms in the expansion of

$(x + a)^{n}$ are

$A$ and

$B$ respectively, then the value of

$(x^{2} - a^{2})^{n}$ is

0%
$A^{2} - B^{2}$
0%
$A^{2} + B^{2}$
0%
$4AB$
0%
None

Explanation

$(x + a)^{n} = ^{n}C_{0} x^{n} + ^{n}C_{1} x^{n - 1}a + ^{n}C_{2} x^{n - 2}a^{2} + ^{n}C_{3} x^{n - 3} a^{3} + ^{n}C_{4}x^{n - 4} a^{4} + .....$

$= (^{n}C_{0}x^{n} + ^{n}C_{2}x^{n - 2}a^{2} + ^{n}C_{4} x^{n - 4} a^{4} + .....) + (^{n}C_{1}x^{n - 1}a + ^{n}C_{3} x^{n - 3}a^{3} + ^{n}C_{5} x^{n - 5}a^{5}) + ....$

$= A + B$ .... (1)
Similarly,

$(x - a)^{n} = A - B$ .... (2)
Multiplying eqns. (1) and (2), we get

$(x^{2} - a^{2})^{n} = A^{2} - B^{2}$

${ _{ }^{ 15 }{ C } }_{ 3 }+{ _{ }^{ 15 }{ C } }_{ 5 }+....+{ _{ }^{ 15 }{ C } }_{ 15 }$ will be equal to

0%
${2}^{14}$
0%
${2}^{14}-15$
0%
${2}^{14}+15$
0%
${2}^{14}-1$

Explanation

We know

${ _{ }^{ 15 }{ C } }_{ 1 }+{ _{ }^{ 15 }{ C } }_{ 3 }+{ _{ }^{ 15 }{ C } }_{ 5 }+.....{ _{ }^{ 15 }{ C } }_{ 15 }={ 2 }^{ 15-1 }$

$\therefore { _{ }^{ 15 }{ C } }_{ 3 }+{ _{ }^{ 15 }{ C } }_{ 5 }+...+{ _{ }^{ 15 }{ C } }_{ 15 }={ 2 }^{ 14 }-15$

Find the sum of coefficient of middle terms of the expansion

$\left(3x-\dfrac{x^3}{6}\right)^7$ :

0%
$\dfrac {595}{48}$
0%
$-\dfrac {595}{48}$
0%
$-\dfrac {595}{24}$
0%
None of the above

Explanation

Total number of terms are

$8$ .
So, middle term will be the

$4^{th}$ and

$5^{th}$ term.

$\therefore t_{3+1}= ^{7}C_3 (-1)^3 (3x)^{7-3} \left(\dfrac {x^3}{6}\right)^3=-\dfrac{105x^{13}}{8}$

$\therefore t_{4+1}= ^{7}C_4 (-1)^4 (3x)^{7-4} \left(\dfrac {x^3}{6}\right)^4=\dfrac{35x^{15}}{48}$
So,

$-\dfrac {105}{8}+\dfrac{35}{48}=-\dfrac {595}{48}$
Ans-Option

$A$ .

The total number of terms in the expansion of

$(x+y)^{50}+(x-y)^{50}$ is

0%
$51$
0%
$26$
0%
$102$
0%
$25$

Explanation

Consider given the expression,

${{\left( x+y \right)}^{50}}+{{\left( x-y \right)}^{50}}$

We know that,

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

Now,

${{\left( x+y \right)}^{50}}+{{\left( x-y \right)}^{50}}=2{{[}^{50}}{{C}_{0}}{{x}^{50}}{{+}^{50}}{{C}_{2}}{{x}^{50-2}}y{{+}^{50}}{{C}_{4}}{{x}^{50-4}}{{y}^{4}}+{{......}^{50}}{{C}_{r}}{{x}^{50-r}}{{y}^{r}}{{.....}^{50}}{{C}_{50}}{{y}^{50}}$

Hence, there are $26$ terms

Hence, this is the answer.

In the expansion of

$(\sqrt 2+\sqrt [3]{5})^{20}$ the number of rational terms will be:

0%
$3$
0%
$10$
0%
$4$
0%
$8$

Explanation

Given

$(\sqrt{2}+\sqrt[3]{5})^{20}$

General term of given expamsion

$T_{r+1}=\displaystyle^{20}C_{r}(\sqrt{2})^{20-r}(\sqrt[3]{5})^r$

$T_{r+1}=\displaystyle^{20}C_{r}(2)^{\dfrac{20-r}{2}}(5)^{\dfrac{r}{3}}$

Now,

$T_{r+1}$ be rational if

$\dfrac{20-r}{2}$ and

$\dfrac{r}{3}$ are integers

Hence

$20-r$ Should be multiple of

$2$ and

$r$ should be multiple of

$3$ where

$0\le r\le 20$

Hence possible value of

$r=0,3,6,9,12,15,18$

$20-r=20,17,14,11,8,5,2$

$20-r$ should be multiple of

$2$

Hence

$20-r=20,14,8,2$

Hence possible rational terms are

$4$

What is

$\displaystyle\sum _{ r=0 }^{ n }{ C\left( n,r \right) }$ equal to?

0%
${2}^{n} - 1$
0%
$n$
0%
$n$ !
0%
${2}^{n}$

Explanation

Now by binomial theorem,

$(1+x)^{ n }={}^{ n }C_{ 0 }+{}^{n}C_{ 1 } x+{}^{ n }C_{ 2 } x^{2}+.........+{}^{ n }C_{ n } x^{n}$

Putting

$x=1$ here we have,

$\Rightarrow 2^{ n }=\displaystyle \sum _{ r=0 }^{ n }{ C(n,r) }$

Hence, option D is correct.

The number of rational terms in the expansion of

$(9^{1/4} + 8^{1/6})^{1000}$ is:

0%
$500$
0%
$400$
0%
$501$
0%
none of the above

Explanation

The general term in the expansion of

$(9^{1/4} + 8^{1/6})^{1000}$ is

$T_{r+1}={}^{1000}C_r\left (9^{\frac {1}{4}}\right )^{1000-r}\left (8^{\frac {1}{6}}\right )^r$

$={}^{1000}C_r 3^{\frac {1000-r}{2}} 2^{\frac {1}{2}}$
The above term will be rational if exponent of 3 and 2 are integers.
It means

$\dfrac {1000-r}{2}$ and

$\dfrac {r}{2}$ must be integers
The possible set of values of

$r$ is

$\left \{0, 2, 4, ...., 1000\right \}$
Hence, number of rational terms is

$501$ .

The

$3rd$ term of

${\left( {3x - \dfrac{{{y^3}}}{6}} \right)^4}$ is

0%
$^4{C_2}{\left( {3x} \right)^2}{\left( { - \dfrac{{{y^3}}}{6}} \right)^2}$
0%
$^4{C_2}{\left( {3x} \right)^2}{\left( { - \dfrac{{{y^2}}}{6}} \right)^2}$
0%
$^4{C_2}{\left( {3x} \right)^3}{\left( { - \dfrac{{{y^3}}}{6}} \right)^2}$
0%
$^4{C_2}{\left( {3x} \right)^3}{\left( { - \dfrac{{{y^2}}}{6}} \right)^2}$

Explanation

As we know the binomial expansion-

$(a+b)^n=^nC_0a^nb^0+^nC_1a^{n-1}b^1+.......+^nC_na^0b^n$

Similarly,

$\left(3x-\dfrac{y^3}{6}\right)^4=^4C_0(3x)^4\left(-\dfrac{y^3}{6}\right)^0+^4C_1(3x)^3\left(-\dfrac{y^3}{6}\right)^1+^4C_2(3x)^2(-\dfrac{y^3}{6})^2+............$

So, the

$3^{rd}$ term

$=^4C_2(3x)^2(-\dfrac{y^3}{6})^2$

[AS 1] If

$A = \dfrac{1}{3} B \, and \, B = \dfrac{1}{2} C$ , then A : B : C = ..

0%
1 : 3 : 6
0%
2 : 3 : 6
0%
3 : 2 : 6
0%
3 : 1 : 2

Explanation

$A=\frac{B}{3}$ ........(1)

$B=\frac{C}{2}$

$\Rightarrow C=2B$ .........(2)

From (1) and (2),

$A:B:C=\frac{B}{3}:B:2B$

$=\frac{1}{3}:1:2$

$=1:3:6$

The expression

$^{n}\textrm{C}_{0}+4\ ^{n}\textrm{C}_{1}+4^{2}\ ^{n}\textrm{C}_{2}+..........+4^{n}\ ^{n}\textrm{C}_{n}$ , equals

0%
$2^{2n}$
0%
$2^{3n}$
0%
$5^{n}$
0%
None of these

Explanation

We know that

$\begin{array}{l} ^{ n }{ C_{ 0 } }+{ 4^{ n } }{ C_{ 1 } }+{ 4^{ 2 } }^{ n }{ C_{ 2 } }+.......+{ 4^{ n } }\, \, { \, ^{ n } }{ C_{ n } } \\ { \left( { 1+x } \right) ^{ n } }{ =^{ n } }{ C_{ 0 } }{ +^{ n } }{ C_{ 1 } }x{ +^{ n } }{ C_{ 2 } }{ x^{ 2 } }+.....{ +^{ n } }{ C_{ n } } \\ Putting\, x=4 \\ { 5^{ n } }{ =^{ n } }{ C_{ 0 } }+{ 4^{ n } }{ C_{ 1 } }+{ 4^{ 2 } }^{ n }{ C_{ 2 } }+.......+{ 4^{ n } }\, \, \, { \, ^{ n } }{ C_{ n } } \\ Hence,\, option\, C\, is\, the\, correct\, answer. \end{array}$

The number of zeroes at the end of

$(101)^{11}-1$ is

0%
$8$
0%
$4$
0%
$1100$
0%
$2$

The middle terms in the expansion of

$(x^{2}-a^{2})^{5}$ is

0%
$10x^{6}a^{4},\ -10x^{4}a^{6}$
0%
$-10x^{6}a^{4},\ 10x^{4}a^{6}$
0%
$10x^{6}a^{4},\ 10x^{4}a^{6}$
0%
$-10x^{6}a^{4},\ -10x^{4}a^{6}$

Explanation

$(a\pm b)^{n} =^{n}C_{0}a^{n}b^{0}\pm ^{n}C_{1}a^{n-1}b^{1}+^{n}C_{2}a^{n-2}b^{2}\pm ....+^{n}C_{n}a^{0}b^{n}$

$(x^{2}-a^{2})^{5} = ^{5}C_{0}x^{10}-^{5}C_{1}x^{8}a^{2}+^{5}C_{2}x^{6}a^{4}-^{5}C_{3}x^{4}a^{6}+^{5}C_{4}x^{2}a^{8}-^{5}C_{5}x^{0}a^{10}$

$= x^{10} - 5x^{8}a^{2}+10x^{6}a^{4} - 10x^{4}a^{6} +5x^{2}a^{8} - x^{0}a^{10}$

Middle term =

$10x^{6}a^{4},-10x^{4}a^{6}$

1229450_1298531_ans_7625ee0915024fdcad0b554098ee038e.jpg

If sum of the coefficients in the expansion of

$(2+3cx+c^2x^2)^{12}$ vanishes, then

$c$ equals to

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

Explanation

sum of the coefficients in the expansion of

$(2+3cx+c^2x^2)^{12}$ is obtained by putting

$x=1$ in the given expression.

Since the sum of the co-efficients vanishes then,

$2+3c+c^2=0$

or,

$(c+1)(c+2)=0$

or,

$c=-1,-2$ .

$^{100}C_{0}-^{100}C_{2}+^{100}C_{4}+^{100}C_{8}-........+^{100}C_{100}=$ __

0%
$2^{-49}$
0%
$-2$
0%
$2^{50}$
0%
$-2^{50}$

If the sum of binomial coefficient in the expansion

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

$256$ , then

$n$ is

0%
$6$
0%
$7$
0%
$8$
0%
$9$

Explanation

$(1+x)^n=C_0+C_1x+\cdots+C_nx^n$

Sum of coefficients is

$C_0+C_1+\cdots+C_n=2^n\\C_0+C_1+\cdots+C_n=256\\2^n=256\\2^n=2^8\\n=8$

The coefficient of

${ x }^{ 3 }$ in the polynomial

$(x-1) (x-2) (x-4)$ is

0%
1
0%
-1
0%
7
0%
None of these

If the constants term in the expansions of

$(\sqrt{x}-\dfrac{k}{x^2})^{10}$ is

$405$ , then what can be the value of

$k$ ?

0%
$\pm 2$
0%
$\pm 3$
0%
$\pm 5$
0%
$\pm 9$

Explanation

Given,

$\left ( \sqrt{x}-\dfrac{k}{x^2} \right )^{10}$

$T_{r+1}={}^{10}C_r(\sqrt{x})^{10-r}\left ( -\dfrac{k}{x^2} \right )^r$

$=^{10}C_r(x)^{\frac{1}{2}(10-r)}(-k)^rx^{-2r}$

$=^{10}C_r(x)^{\frac {10-5r}{2}}(-k)^r$

For a constant term, the condition is

$=x^0$

$\Rightarrow \dfrac {10-5r}{2}=0$

$\Rightarrow r=2$

$T_{2+1}=^{10}C_2(-k)^2$

Given,

$405=^{10}C_2(-k)^2$

$\dfrac{10 \times 9 \times 8!}{2! \times 8!}(-k)^2=405$

$45k^2=405$

$k^2=9\Rightarrow k= \pm 3$

The middle term in

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

0%
$\dfrac{{n!}}{{{{\left( {\left( {\frac{n}{2}} \right)!} \right)}^2}}}$
0%
$\dfrac{{\left( {2n} \right)!}}{{{{\left( {\left( {\frac{n}{2}} \right)!} \right)}^2}}}$
0%
$\dfrac{{1.3.5....\left( {2n + 1} \right)}}{{n!}}{2^n}$
0%
$\dfrac{{\left( {2n} \right)!}}{{{{\left( {n!} \right)}^2}}}$

Explanation

$(x^2+\cfrac{1}{x^2}+2)^n=(x+\cfrac{1}{x})^{2n}$

Middle term

$=(n+1)th \quad term$

$T_{n+1}= { { ^{ 2n }{ C } } }_{ n }(x)^n(\cfrac{1}{x})^n\\ \quad=\cfrac{2n!}{(n!)^2}$

The term containing

$x^3$ in the expansion of

$(x - 2y)^7$ is

0%
$3$ rd
0%
$4$ th
0%
$5$ th
0%
$6$ th

Explanation

Use the formula

$T_{r+1} = {(-1)}^r \; ^nC_r x^{n-r} a^r$

For

$(r + 1)$ th term and putting

$n = 7$ ,

we get

$7-r = 3$ for power

$3$ term of

$x.$

Hence,

$r = 4$ and

$r+1 = 5$

So, it's

$5$ th term.

The greatest coefficient in the expansion of

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

0%
$\frac{\left ( 2n \right )!}{\left ( n! \right )^{2}}$
0%
$\frac{\left ( 2n+2 \right )!}{\left (( n+1 \right )!)^{2}}$
0%
$\frac{\left ( 2n+2 \right )!}{n!\left (n+1 \right )!}$
0%
$\frac{\left ( 2n \right )!}{n!\left (n+1 \right )!}$

In the expansion of

$(2+\dfrac{x}{3})^n$ , coefficients of

$x^7$ and

$x^8$ are equal. Then

$n =$

0%
$49$
0%
$50$
0%
$55$
0%
$56$

Explanation

We know,

$T_{r+1}=\:^nC_{r}a^{n-r}b^r$
Applying to the above question, we get

$T_{r+1}=\:^nC_{r}2^{n-r}3^{-r}x^{r}$
Applying the given conditions, i.e. coefficient of

$x^7$ is equal to coefficient of

$x^8$
Therefore

$\:^nC_{7}2^{n-7}3^{-7}=\:^nC_{8}2^{n-8}3^{-8}$

$\Rightarrow 6\:^nC_{7}=\:^nC_{8}$

$\Rightarrow \dfrac{6.(n)!}{(n-7)!7!}=\dfrac{(n)!}{(n-8)!8!}$

$\Rightarrow 48=n-7$

$\Rightarrow n =55$

The middle term in the expansion of

$\left (\displaystyle \frac{x^{\frac{3}{2}}}{\sqrt{a}} - \frac{y^{\frac{5}{2}}}{b^{\frac{3}{2}}} \right )^8$ is

0%
$^8C_4. \dfrac{x^6.y^{10}}{a^2b^6}$
0%
$^8C_4. \dfrac{x^6.y^{8}}{a^2b^4}$
0%
$^8C_5. \dfrac{x^5.y^{10}}{a^2b^5}$
0%
$^8C_1. \dfrac{x^5.y^{10}}{a^2b^5}$

The middle term in the expansion of

$\left ( x + \dfrac{1}{x} \right )^{10}$ is

0%
$^{10}C_{4}.\dfrac{1}{x}$
0%
$^{10}C_{5}$
0%
$^{10}C_{5}.\dfrac{1}{x}$
0%
$^{10}C_{6}.x$

Explanation

Middle term

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

$=6^{th}$ term
Now consider the following:

$T_{r+1}=\:^nC_{r}a^{n-r}b^r$ ...(i)
Where

$T$ represents the term.
Here the middle term is the

$6^{th}$ term.

Hence

$r+1=6$

$\therefore r=5$
substituting in (i), we get

$\:^{10}C_{5}x^{10-5}\frac{1}{x^5}$

$=\:^{10}C_{5}x^{5}x^{-5}$

$=\:^{10}C_{5}$
Hence, answer is B.

The middle term in the expansion of

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

0%
$^{2n}C_{n}$
0%
$^{2n}C_{n-1}.x^{n+1}$
0%
$^{2n}C_{n-1}.x^{n-1}$
0%
$^{2n}C_{n}.x^n$

Explanation

Middle term

$=\dfrac{N}{2}+1$

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

$=(n+1)^{th}$ term
Now consider the following,

$T_{r+1}=\:^nC_{r}a^{n-r}b^r$ ...(i)
Where

$T$ represents the term
Here the middle term is the

$(n+1)^{th}$ term
Hence

$r+1=n+1$

$\therefore r=n$
Substituting in (i), we get

$\:^{2n}C_{n}1^{2n-n}x^{n}$

$=\:^{2n}C_{n}x^{n}$
Hence answer is D

The middle term of

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

0%
$^{2n+1}C_n.x$
0%
$^{2n+1}C_n$
0%
$(-1)^{n}.^{2n+1}C_{n}$
0%
$(-1)^{n}.^{2n+1}C_{n}.x$

Explanation

Let the middle term of

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

$M$ .

The middle term is

$\dfrac{1}{2}(2n+1+1)$ i.e

$(n+1)^{th}$ term.

$M=T_{n+1}$

We k.n.t

$T_r$ of expansion

$(a+b)^n$ is

$T_r=^nC_{r-1}a^{n-r+1}b^{r-1}$

$\therefore M=T_{n+1}=^{2n+1}C_n x^{2n+1-n} \left (-\dfrac{1}{x}\right)^n = (-1)^n .^{2n+1}C_n.x^{n+1-n}=(-1)^n.^{2n+1}C_n.x$

Option (D) is correct.

If the coefficients of

$(r+2)^{th}$ and

$(2r+1) ^{th}$ terms

$(r \neq 1)$ are equal in the expansion of

$(1+x)^{43}$ , then

$r =$

0%
$12$
0%
$13$
0%
$14$
0%
$15$

Explanation

Since coefficient of

$(r+2)^{th}$ term is equal to coefficient of

$(2r+1)^{th}$ term in the binomial expansion of

$(1+x)^{43}$
Therefore

$\:^{43}C_{r+1}=\:^{43}C_{2r}$
Now

$\:^{n}C_{r}=\:^{n}C_{n-r}$

$\therefore 43-(r+1)=2r$

$\Rightarrow 42=3r$

$\Rightarrow r=14$

Sum of the coefficients of

$(1+x)^n$ is always a

0%
an integer
0%
positive integer
0%
negative integer
0%
zero

Explanation

To determine the sum of coefficients, we substitute

$x=1$ in the above expression.

Thus sum of coefficients

$=(1+1)^{n}$

$=2^{n}$

Hence, its a positive integer.

The number of terms in the expansion of

$(1+x)^{21}$ is

0%
$20$
0%
$21$
0%
$22$
0%
$24$

Explanation

$\textbf{Step 1: Find the total number of terms in a binomial expansion.}$

$\text{The number of terms in the binomial expansion is one more than power }n$

$\text{i.e., } n + 1$

$\text{So, here }n = 21$

$\text{The number of terms in the expansion }(1+x)^{21}=21 + 1 = 22$ .

$\textbf{Hence, Option C is correct.}$

The ratio of the

${ r }^{ th }$ term and the

$({ r+1 })^{ th }$ term in the expansion of

$(1+ x)^n$ is

0%
$\displaystyle \frac{r}{(n-r+1)x}$
0%
$\displaystyle \frac{1}{(n-r+1)x}$
0%
$\displaystyle \frac{r}{(n-r+1)}$
0%
$\displaystyle \frac{(n-r+1)x}{r}$

Explanation

$T_{r}^{th}$ term

$=\:^{n}C_{r-1}x^{r-1}$ ...(i)
And

$T_{r+1}$ th term

$=\:^{n}C_{r}x^{r}$ ...(ii)
Therefore,

$\displaystyle \frac{T_{r}}{T_{r+1}}$

$=\displaystyle \frac{\:^{n}C_{r-1}x^{r-1}}{\:^{n}C_{r}x^{r}}$

$=\displaystyle \frac{\:^{n}C_{r-1}}{\:^{n}C_{r}x}$

$=\displaystyle \frac{r}{(n-(r-1))x}$

$=\displaystyle \frac{r}{(n-r+1)x}$

In the binomial expansion of

$(a-b)^n , n \geq{5}$ , the sum of
5th and 6th terms is zero then a/b equal to

0%
$\frac{5}{n-4}$
0%
$\frac{6}{n-5}$
0%
$\frac{n-5}{6}$
0%
$\frac{n-4}{5}$

Explanation

It is given that

$T_{6}+T_{5}=0$ .
Hence

$\:^{n}C_{4}a^{n-4}b^{4}-\:^{n}C_{5}a^{n-5}b^{5}=0$

$\:^{n}C_{4}a^{n-4}b^{4}=\:^{n}C_{5}a^{n-5}b^{5}$

$\:^{n}C_{4}a=\:^{n}C_{5}b$

$\displaystyle \frac{n!a}{(n-4)!4!}=\frac{n!b}{(n-5)!5!}$

$\displaystyle \frac{a}{(n-4)}=\frac{b}{5}$
Therefore

$\displaystyle \frac{a}{b}=\frac{n-4}{5}$

If the middle term of

$(1+x)^{2n}$ is the greatest term then

$x$ lies between

0%
$n - 1 < x < n$
0%
$\dfrac{n}{n + 1} < x < \dfrac{n+1}{n}$
0%
$n < x < n+1$
0%
$\dfrac{n+1}{n} < x < \dfrac{n}{n+1}$

Explanation

The numerically greatest term in

$(1+x)^n$ will be given by.

$\displaystyle \frac{(1+n)(|x|)}{1+|x|}$

Now,
In

$(1+x)^{2n}$
The middle term would be

$\dfrac{2n}{2}+1$ th term

Hence

$T_{n+1}$ would be the middle term.

There applying condition for a numerically greatest term we get.

$n<\dfrac{(2n+1)|x|}{|x|+1}<n+1$ ...(since there would be total of 2n+1 terms).

$=n|x|+n<2n|x|+|x|$

$2n|x|+|x|<n|x|+x+n+1$

$=n<n|x|+|x|$

$n|x|<n+1$

$=n<|x|(n+1)$

$|x|<\dfrac{n+1}{n}$ ...(i)

$=\dfrac{n}{n+1}<|x|$ ...(ii)

Hence from i and ii we get

$\dfrac{n}{n+1}<|x|<\dfrac{n+1}{n}$

Hence answer is Option B

In the expansion of

$(a+b)^n$ , the ratio of the binomial coefficients of

${ 2 }^{ nd }$ and

${ 3 }^{ rd }$ terms is equal to the ratio of the binomial coefficients of

${ 5 }^{ th }$ and

${ 4 }^{ th }$ terms, then

$n =$

0%
$4$
0%
$5$
0%
$6$
0%
$7$

Explanation

It is given that coefficients of

$\displaystyle \frac{T_{2}}{T_{3}}=\frac{T_{5}}{T_{4}}$
Therefore,

$\displaystyle \frac{\:^nC_{1}}{\:^nC_{2}}=\frac{\:^nC_{4}}{\:^nC_{3}}$

$\Rightarrow \displaystyle \dfrac{n}{\frac{n(n-1)}{2!}}=\frac{\frac{(n)(n-1)(n-2)(n-3)}{4!}}{\frac{n(n-1)(n-2)}{3!}}$

$\Rightarrow \displaystyle \frac{2}{n-1}=\frac{n-3}{4}$

$8=n^2-4n+3$

$n^2-4n-5=0$

$(n-5)(n+1)=0$

$n=5$ and

$n=-1$
However,

$n$ is a natural number.
Therefore

$n=5$

The ratio of (r+1)th and rth terms in the expansion of

$(1-x)^n$ is

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

Explanation

$T_{r}$ th term

$=(-1)^{r-1}\:^{n}C_{r-1}x^{r-1}$ ...(i)
And

$T_{r+1}$ th term

$=(-1)^{r}\:^{n}C_{r}x^{r}$ ...(ii)
Therefore

$\displaystyle \frac{T_{r+1}}{T_{r}}$

$=\displaystyle \frac{-\:^{n}C_{r}x^{r}}{\:^{n}C_{r-1}x^{r-1}}$

$=\displaystyle \frac{-\:^{n}C_{r}x}{\:^{n}C_{r-1}}$

$=-\displaystyle \frac{(n-(r-1))x}{r}$

$=-\displaystyle \frac{(n-r+1)x}{r}$

In the expansion of

$\left ( a^2\sqrt{a} + \frac{\sqrt[3]{a}}{a} \right )^n$ the binomial coefficient of 3rd term isThe 7th term is :

0%
$84a^3\sqrt{a}$
0%
$84a^2\sqrt{a}$
0%
$84 a^2$
0%
$84 a^3$

Explanation

It is given that :

$\:^nC_{2}=36$

$\frac{n(n-1)}{2!}=36$

$n(n-1)=72$

$n(n-1)=9(8)$
Hence

$n=9$
Hence

$T_{7}=\:^9C_{6}(a)^{\frac{15}{2}}a^{-4}$

$=\:^9C_{3}a^{\frac{7}{2}}$

$=\dfrac{9.8.7}{3!}.(a^3\sqrt{a})$

$=84(a^3\sqrt{a})$

In the expansion of

$\displaystyle \left ( \sqrt{a} + \frac{1}{\sqrt{(3a)}} \right )^n$ , if the ratio of the binomial coefficient of the

$4^{th}$ term to the binomial coefficient of the

$3^{rd}$ term is

$\dfrac{10}{3}$ , the

$5^{th}$ term is

0%
$\dfrac{88a}{\sqrt 3}$
0%
$\dfrac {88a}{3}$
0%
$50a^2$
0%
$55a^2$

Explanation

The ratio of the binomial coefficient of

$T_{4}$ to that of

$T_{3}$ will be:

$\displaystyle \frac{\:^nC_{3}}{\:^nC_{2}}$

$=\displaystyle \frac{\frac{(n)(n-1)(n-2)}{6}}{\frac{n(n-1)}{2}}$

$=\displaystyle \frac{n-2}{3}$

$=\displaystyle \frac{10}{3}$ ...(given)
Hence,

$n=12$ .
Therefore,

$T_{5}$

$=\:^{12}C_{4}a^4\displaystyle \frac{1}{9a^2}$

$=\displaystyle \frac{12.11.10.9}{4!}a^4\frac{1}{9a^2}$

$=55a^2$

Sum of the coefficients of

$(1 - x)^{25}$ is

0%
$-1$
0%
$1$
0%
$0$
0%
$2^{25}$

Explanation

$(1-x)\:^{25}$

$=1-\:^{25}C_{1}x+\:^{25}C_{2}x^2-\:^{25}C_{3}x^3+\:^{25}C_{4}x^4-\:^{25}C_{5}x^5...-\:^{25}C_{25}x^{25}$
Putting

$x=1$ , we get

$0=1-\:^{25}C_{1}+\:^{25}C_{2}-\:^{25}C_{3}+\:^{25}C_{4}-\:^{25}C_{5}...-\:^{25}C_{25}$
Hence, sum of coefficients is

$0$

The product of two middle terms in the expansion of

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

0%
$(^9C_4)^2.\displaystyle \frac{x^9}{512}$
0%
$^{-9}C_4.^9C_5.\displaystyle \frac{x^8}{512}$
0%
$-^{9}C_4.^9C_5.\displaystyle \frac{x^9}{512}$
0%
$^9C_4.^9C_5.\displaystyle \frac{x^9}{256}$

Explanation

Since 9 is odd the middle terms are

$(\dfrac{N+1}{2})$ th and

$(\dfrac{N+1}{2}+1)$ th terms
Here the middle terms are the 5th term and 6th term.
Hence

$r+1=5$ and

$r+1=6$

$r=4$ and

$r=5$

$T_{4+1}=\:^9C_{4}(\dfrac{3x^2}{2})\:^{9-4}(\dfrac{-1}{3x})\:^4$

$=\:^9C_{4}(\dfrac{3x^2}{2})\:^{5}(\dfrac{1}{3x})\:^4$ ...(i)

$T_{5+1}=\:^9C_{5}(\dfrac{3x^2}{2})\:^{9-5}(\dfrac{-1}{3x})\:^5$

$=-\:^9C_{5}(\dfrac{3x^2}{2})\:^{4}(\dfrac{1}{3x})\:^5$ ...(ii)

Multiplying i and ii we get

$[\:^9C_{4}(\dfrac{3x^2}{2})\:^{5}(\dfrac{1}{3x})\:^4][-\:^9C_{5}(\dfrac{3x^2}{2})\:^{4}(\dfrac{1}{3x})\:^5]$

$=-\:^9C_{4}\:^9C_{5}\dfrac{x^9}{2^9}$

$=-\:^9C_{4}\:^9C_{5}\dfrac{x^9}{512}$

The

$3$ rd,

$4$ th and

$5$ th terms in the expansion of

$(1+x)^n$ are

$60, 160$ and

$240$ respectively, then

$x =$

0%
$2$
0%
$4$
0%
$5$
0%
$6$

Explanation

It is given that in

$(1+x)^{n}$

$T_{3}=\:^nC_{2}x^2=60$

$T_{4}=\:^nC_{3}x^3=160$

$T_{5}=\:^nC_{4}x^4=240$
Therefore taking ratios we get

$\displaystyle \frac{\:^nC_{2}x^2}{\:^nC_{3}x^3}=\frac{3}{8}$

$\displaystyle \frac{3}{x(n-2)}=\frac{3}{8}$

$8=nx-2x$

$nx=2x+8$

$...(i)$

$\displaystyle \frac{\:^nC_{3}x^3}{\:^nC_{4}x^4}=\frac{160}{240}$

$\displaystyle \frac{4}{x(n-3)}=\frac{160}{240}$

$\displaystyle \frac{1}{nx-3x}=\frac{40}{240}$

$6=nx-3x$

$nx=3x+6$

$...(ii)$
Substituting the value of

$nx$ in equation

$(i)$ we get

$3x+6=2x+8$

$x=2$

$\displaystyle \frac{^{15}C_1}{^{15}C_0}+2.\frac{^{15}C_2}{^{15}C_1}+3.\frac{^{15}C_3}{^{15}C_2}+\ldots+15.\frac{^{15}C_{15}}{^{15}C_{14}} =$

0%
105
0%
91
0%
120
0%
15

Explanation

As in the hint , This is the first step. Now we have to find sum of

$r(n - r + 1)/r$ = sum of

$(n - r + 1)$ =

$n*n - n(n+1)/2 + n = n(n+1)/2$ . Here n= 15 which gives answer 120.

The number of terms in the expansion of

$(1+5\sqrt{2}x)^9 + (1-5\sqrt{2}x)^9$ is :

0%
$5$
0%
$10$
0%
$18$
0%
$20$

Explanation

$(1+5\sqrt{2}x)\:^{9}+(1-5\sqrt{2}x)\:^{9}$
Applying binomial theorem for expansion of the following

$=1+(5\sqrt{2}x)\:(^{9}C_{1})+(5\sqrt{2}x)^{2}\:(^{9}C_{2})...(\:^{9}C_{9}(5\sqrt{2}x)^{9})$

$+1-(5\sqrt{2}x)\:(^{9}C_{1})+(5\sqrt{2}x)^{2}\:(^{9}C_{2})...-(\:^{9}C_{9}(5\sqrt{2}x)^{9})$

$=2[1+(5\sqrt{2}x)\:^2\:(^{9}C_{2})+(5\sqrt{2}x)\:^4\:(^{9}C_{4})...(5\sqrt{2}x)\:^8(\:^{9}C_{8})]$
The powers of

$5\sqrt{2}x$ are in

$A.P$

$0,2,4,6,8$
Hence including the term

$1\:or\: x^0$ there are

$\dfrac{9+1}{2} = 5$ terms.
Hence, answer is

$A.$

$^{2n}C_n = C_0^2 + C_1^2 + C_2^2 + C_3^2 + \dots \dots + C_n^2$

B.

$^{2n}C_n =$ term independent of

$x$ in

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

C.

$^{2n}C_n = \dfrac{1.3.5.7 \ldots \ldots (2n-1)}{n!}$ then

0%
A, B are false, C is true
0%
A is false, B and C are true
0%
A and B are true; C is false
0%
A, B, C are true

Explanation

$\:^{2n}C_{n}$

$=\dfrac{2n!}{(n!)^{2}}$

$=\dfrac{2^{n}(1.3.5..(2n-1))}{n!}$

Hence option C is false.

Consider Option B

$(1+x)^{n}\cdot (1+\dfrac{1}{x})^{n}$

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

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

The general term in this case is

$T_{r+1}=\:^{2n}C_{r}x^{r}.x^{-n}$

$=\:^{2n}C_{r}x^{r-n}$
For term independent of

$x$

$r-n=0$

$n=r$ .

Hence the term independent of

$x$ will be

$\:^{2n}C_{n}$
Thus B is true.

Consider option A.
Consider the following series.

$\:^{m}C_{0}\:^{n}C_{r}+\:^{m}C_{1}\:^{n}C_{r-1}+...\:^{m}C_{r}\:^{n}C_{0}$

$=$ Coefficient of

$x^{r}$ in

$(1+x)^{m}.(x+1)^{n}$

$=$ Coefficient of

$x^{r}$ in

$(1+x)^{m+n}$

$=\:^{m+n}C_{r}$

In the above case,

$r=n$ and

$m=n$
Hence,

$\:^{n}C_{0} ^{2}+\:^{n}C_{1} ^{2}+\:^{n}C_{2} ^{2}+\:^{n}C_{3} ^{2}+..\:^{n}C_{n} ^{2}$

$=\:^{n}C_{0}\:^{n}C_{n}+\:^{n}C_{1}\:^{n}C_{n-1}+...\:^{n}C_{2}\:^{n}C_{n-2}$

$=\:^{2n}C_{n}$

Hence A is also true.

The total number of terms in the expansion of

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

0%
$202$
0%
$51$
0%
$101$
0%
$50$

Explanation

$(x+a)\:^{100}+(x-a)\:^{100}$
Applying binomial theorem for expansion of the following

$(x+a)\:^{100}+(x-a)\:^{100}$

$=x^{100}+x^{99}\:(^{100}C_{1}a)+x^{98}\:(^{100}C_{2}a^2)...(\:^{100}C_{100}a^{100})$

$+x^{100}-x^{99}\:(^{100}C_{1}a)+x^{98}\:(^{100}C_{2}a^2)...(\:^{100}C_{100}a^{100})$

$=2[x^{100}+x^{98}\:(^{100}C_{2}a^2)+x^{96}\:(^{100}C_{4}a^4)...(\:^{100}C_{100}a^{100})]$
The powers of

$a$ are in A.P

$a^0,a^2,a^4,a^6...a^{100}$

$0,2,4,6...100$
Hence, including the term

$a^0\:or\:x^{100}$ there are

$51$ terms.
Hence, answer is

$B.$

$C_0^2+3.C_1^2+5.C_2^2 + \ldots\ldots +(2n+1).C_n^2 =$

0%
$(n+1)2^n$
0%
$(2n+1) ^{2n}C_n$
0%
$(n+1). ^{2n}C_n$
0%
$(2n-1) ^{2n}C_n$

Explanation

$(1+x)^n=C_0+C_1 x+C_2 x^2+\dots +C_n x^n$ ------(1)

$\displaystyle(1+\frac{1}{x})^n=C_0+C_1\frac{1}{x}+C_2\frac{1}{x^2}+\dots +C_n\frac{1}{x^n}$ -----(2)
Multiplying both sides, we get

$\displaystyle\frac{(1+x)^{2n}}{(x)^n}=\sum{C_0^{2}}+x\sum{C_0C_1}+x^2\sum{C_0C_2}+ ....+x^r\sum{C_0C_r}+...$
The various sums are the coefficients of

$x^0,x,x^2,....x^r$ in

$\displaystyle\frac{(1+x)^{2n}}{(x)^n}$ or coefficients of

$x^n,x^{n+1},x^{n+2},....,x^{n+r}$ in the expansion of

$(1+x)^{2n}$ which occur in

$T_{n+1},T_{n+2}....$
Clearly the coefficent of

$x^n$ in

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

$^{2n}C_n$

$\therefore C_0^2+C_1^2+C_2^2 + \ldots\ldots +C_n^2 = ^{2n}C_n$
Let

$S= C_0^2+3.C_1^2+5.C_2^2 + \ldots\ldots +(2n+1).C_n^2$ ----(3)

$S=(2n+1).C_n^2+(2n-1).C_{n-1}^2+ \ldots\ldots+C_0^2$ -----(4)
Adding (3) and (4) gives

$2S=(2n+2)\left[C_0^2+C_1^2+C_2^2 + \ldots\ldots +C_n^2\right]$

$\therefore S= (n+1). ^{2n}C_n$
Hence, option C.

$^nC_0 + ^nC_2 + ^nC_4 + \dots \dots + ^nC_{2[n/2]}$ , where [ ] denotes greatest integer

0%
$2^{2n-1}$
0%
$2^{2n-1}-1$
0%
$2^{n-1}$
0%
$2^{n-1}-1$

Explanation

The real sol. is taking a binomial =

$(1+x)^{n} = ^{n}C_0 + ^{n}C_1x + ^{n}C_2x^2 + ......$
and now taking

$(1-x)^{n} = ^{n}C_0 - ^{n}C_1x + ^{n}C_2x^2 - ......$
Now adding both and putting

$x = 1$ , we get

$2^n = 2( ^{n}C_0 + ^{n}C_2 + ^{n}C_4 + ....)$
=>

$2^{n-1}$ = required expression

$^5C_0+2.^5C_1+2^2.^5C_2+2^3.^5C_3 +2^4.^5C_4+2^5.^5C_5 =$

0%
32
0%
243
0%
64
0%
729

Explanation

Just calculate it by substituting the values of

$^5C_1 , ^5C_2$ etc as

$5, 10$ and just add them to get

$243$ .

Evaluate the following:

$C_1+2C_2+3C_3+\ldots\dots+nC_n$

0%
$n2^n$
0%
$n2^{n-1}$
0%
$(n+1)2^n$
0%
$(n+1)2^{n-1}$

Explanation

Consider the expansion

$(1 + x)^n = \ ^n C_0 + \ ^nC_1 x + \ ^nC_2 x^2 ..... + \ ^nC_n x^n$

derivate w.r.t 'x'

$n(1 + x)^{n-1} = 0 + \ ^nC_1 + 2 \ ^nC_2 x .... + n \,{}^nC_n x^{n - 1}$

Put

$x = 1$

$n(1 + 1)^{n- 1} = \ ^nC_1 + 2 . \ ^nC_2 ... + n . \ ^nC_n$

$n2 ^{n - 1} = C_1 + 2 C_2 .... + nC_n$

$(1+x)^{15}=a_0+a_1x+\ldots\ldots+a_{15}x^{15} \Rightarrow \sum_{r=1}^{15}r\frac{a_r}{a_{r-1}}=$

0%
110
0%
115
0%
120
0%
135

Explanation

$^nC_r/^nC_{r-1} = n -r + 1/r$ . Now we have to find sum of r(n - r + 1)/r = sum of(n - r + 1) =
n*n - n(n+1)/2 + n = n(n+1)/2. Here n= 15 which gives answer 120.

$n\geq2$ then

$(a-1).C_1-(a-2).C_2+(a-3).C_3-\ldots\ldots(-1)^{n-1}(a-n).C_n=$

0%
$0$
0%
$a-1$
0%
$a$
0%
$a+1$

Explanation

$(a-1)C_{1}-(a-2)C_{2}+(a-3)C_{3}-...(-1)^{n}(a-n)C_{n}$

$=[aC_{1}-aC_{2}+aC_{3}...]-[C_{1}-2C_{2}+C_{3}...]$

$=a(C_{1}-C_{2}+C_{3}...)-(C_{1}-2C_{2}+C_{3}...)$

$=a(1+(-1+C_{1}-C_{2}+C_{3}...))-(C_{1}-2C_{2}+C_{3}...)$

$=a+a(-1+C_{1}-C_{2}+C_{3}...)-(C_{1}-2C_{2}+C_{3}...)$

$=a+-a(1-x)^{n}|_{x=1}-n(1-x)^{n-1}|_{x=1}$

$=a$

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 1 - MCQExams.com

0:0:3

Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 1 - MCQExams.com

0:0:3

Practice Class 11 Engineering Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.