MCQ Questions for Class 11 Engineering Maths Binomial Theorem Quiz 8

$\sum\limits_{k = 1}^{n - r} {{}^{n - k}\mathop C\nolimits_r = {}^x\mathop C\nolimits_y }$

0%
$x = n + 1 ; y = r$
0%
$x = n ; y = r + 1$
0%
$x = n ; y = r$
0%
$x = n + 1 ; y= r + 1$

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

${ C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+{ C }_{ 2 }^{ 2 }+{ C }_{ 3 }^{ 2 }+.......+{ C }_{ n }^{ 2 }$ is equal to

0%
$\dfrac {n!}{n!n!}$
0%
$\dfrac {{(2n)}!}{n!n!}$
0%
$\dfrac {{(2n)}!}{n!}$
0%
$None\ of\ these$

Explanation

$\begin{array}{l} { \left( { 1+x } \right) ^{ h } }={ C_{ 0 } }+{ C_{ 1 } }x+{ C_{ 2 } }{ x^{ 2 } }.........{ e_{ n } }{ x^{ 4 } } \\ { \left( { 1+x } \right) ^{ h } }={ C_{ 0 } }{ x^{ h } }+{ C_{ 1 } }{ x^{ h-1 } } \\ \left( 1 \right) \times \left( 2 \right) \\ { \left( { 1+x } \right) ^{ 2h } }=\left( { { C_{ 0 } }+{ C_{ 1 } }x.....{ C_{ 2 } }{ x^{ 4 } } } \right) \\ comparison \\ 2h{ C_{ n } }=C_{ 0 }^{ 2 }+C_{ 1 }^{ 2 }......C_{ n }^{ 2 } \\ C_{ 0 }^{ 2 }+C_{ 1 }^{ 2 }......C_{ n }^{ 2 }=\frac { { 2n! } }{ { \left( { n!n! } \right) } } \end{array}$

Find the

$13^{th}$ terms in the expansion of

$\left(9x-\dfrac{1}{3\sqrt{x}}\right)^{18}, x \neq 0$ .

0%
$18564$
0%
$87328$
0%
$17374$
0%
$35546$

Explanation

${\left( {9x - \frac{1}{{\sqrt {3x} }}} \right)^{18}}$

${T_{r + 1}} = {}^{18}{C_r}{\left( {9x} \right)^{18 - r}}{\left( { - \dfrac{1}{{3\sqrt x }}} \right)^r}$

putting

$r = 12$

${T_{13}} = {}^{18}{C_{12}}{\left( {9x} \right)^{18 - 12}}{\left( { - \dfrac{1}{{3\sqrt x }}} \right)^r}$

$= {}^{18}{C_6} \times {\left( {9x} \right)^6} \times \dfrac{1}{{{3^{12}} \times {x^6}}}$

$= {}^{18}{C_6} \times {9^6} \times {x^6} \times \dfrac{1}{{{3^{12}} \times {x^6}}}$

$= {}^{18}{C_6} \times {3^{12}} \times \dfrac{1}{{{3^{12}}}}$

$= \dfrac{{18 \times 17 \times 16 \times 15 \times 14 \times 13}}{{6 \times 5 \times 4 \times 3 \times 2}}$

$= 17 \times 4 \times 3 \times 7 \times 13$

$= 18564$

The co-efficient of x in the expansion of

$(1-2x^3 + 3x^5) \left( 1 + \dfrac{1}{x}\right)^8$ is :

0%
56
0%
65
0%
154
0%
62

Explanation

$(1-2x^{3}+3x^{5})(1+\frac{1}{x})^{8}$

$(1-2x^{3}+3x^{5})(^{8}C_{0}+^{8}C_{1}\frac{1}{x}+^{8}C_{2}+^{8}C_{3}\frac{1}{x^{3}}+...^{8}C_{4})$

$= 3.^{8}C_{4}-2^{8}C_{2}$

$= \dfrac{3.8!}{4.4!}-2\times \dfrac{8!}{6!2!}$

$= \dfrac{3\times 8\times 7\times 6\times 5\times 4!}{4\times 3\times 2\times 4!}-\dfrac{2\times 8\times 7\times 6!}{6!\times 2!}$

$= 210-56$

$= 154$

1066886_1070970_ans_88a7311b721f4a6aaf8d10cf5b4bffad.jpg

The third term from the end in the expansion of

$\left( \dfrac{3x}{5}-\dfrac{5}{2x}\right)^8$ is

0%
$\dfrac{35451}{15x^4}$
0%
$\dfrac{45455}{16x^4}$
0%
$\dfrac{39372}{15x^4}$
0%
$\dfrac{39375}{16x^4}$

Explanation

$Third\>term\>from\>the\>end\>will\>be\>\\(8-3+2)th\>term\>from\>beginning\\T_7=^8C_6(\dfrac{3x}{5})^2(\dfrac{-5}{2x})^6\\=(\dfrac{8\times\>7}{2})\times\>(\dfrac{9}{25})\times x^2\times\>(\dfrac{15625}{64\times\>x^6})\\=(\dfrac{39375}{16x^4})$

The co-efficient of

${x^5}$ in the expansion of

${\left( {1 + x} \right)^{21}} + {\left( {1 + x} \right)^{22}} + ........ + {\left( {1 + x} \right)^{30}}$ is:

0%
$^{51}{C_5}$
0%
$^{9}{C_5}$
0%
$^{31}{C_6}{ - ^{21}}{C_6}$
0%
$^{30}{C_5}{ + ^{20}}{C_5}$

Explanation

According to question,

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

$(1+x)^{22}$ +..................+

$(1+x)^{30}$

For , coeff. of

$x^5$ in the expansion, we need to take out coeff. of

$x^5$ from every terms of given series.

i.e;

$^{21}C_5+^{22}C_5+.................+^{30}C_5$

Add & subtract

$^{21}C_6$ in this,

$^{21}C_6+^{21}C_5+^{22}C_5+.................+^{30}C_5-^{21}C_6$

Since, (

$^nC_r+^nC{r-1}=^{n+1}C_r$ )

Therefore,

=>

$^{22}C_6+^{22}C_5+.................+^{30}C_5-^{21}C_6$

=>

$^{23}C_6+^{23}C_5+.................+^{30}C_5-^{21}C_6$

Similarly , we get

=>

$^{30}C_6+^{30}C_5-^{21}C_6$

=>

$^{31}C_6-^{21}C_6$

${S_n} = \sum\limits_{r = 0}^n {\dfrac{1}{{^n{C_r}}}\,\,and\,\,{t_n} = \sum\limits_{r = 0}^n {\dfrac{r}{{^n{C_r}}},\,\,then\,\,\dfrac{{{t_n}}}{{{s_n}}} = } }$

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

In the expansion of

$(1+ax)^n$ ,

$n\in N$ , then the coefficient of x and

$x^2$ are

$8$ and

$24$ respectively. Then?

0%
$a=2, n=4$
0%
$a=4, n=2$
0%
$a=2, n=6$
0%
None of these

Explanation

Here in the expansion of

$(1+x.a)^n$ ,

$n\in N$ , the

$(r+1)^{th}$ term will be

$T_{r+1}=\left (\dfrac {n}{r}\right)a^{n-r}. b^r$

here first term

$a=1$ &

second term

$b=x.a$ , so,

$T_{r+1}=\left (\dfrac {n}{r}\right) (1)^{n-r}. (x.a)^r=\left (\dfrac {n}{r}\right).x^r.a^r$

But given that, coefficient of

$x$ is

$8$ so,

For

$x$ power

$1$ , we have to put

$r=1$

So,

$T_2=\left (\dfrac {n}{1}\right)a^1 .x^1$

So,

$\left (\dfrac {n}{1}\right)a=8\to (i)$

Now, coefficient of

$x^2$ is

$24$ , so,

$r=2$

$T_3=\left (\dfrac {n}{2}\right) a^2 .x^2$

So,

$24=\left (\dfrac {n}{2}\right) a^2\to (ii)$

From

$(1)$ &

$(2), a.n=8\ \Rightarrow \ a=8/n$

$\dfrac {n(n-1)}{2}.a^2=24$

$\dfrac {n(n-1)}{2}\left (\dfrac {8}{n}\right)^2=24$

$\therefore \ \dfrac {n^2-n}{n^2}\times 32=24$

$\therefore \ 1-\dfrac {1}{n}=\dfrac {24}{32}=\dfrac {3}{4}$ , So,

$\boxed {n=4}$ from

$(i)$

$\boxed {a=2}$

$\left| x \right| < 1$ then the coefficient of

${x^n}$ in expansion of

${\left( {1 + x + {x^2} + {x^3}....} \right)^2}$ is

0%
$n$
0%
$n - 1$
0%
$n + 2$
0%
$n + 1$

Explanation

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

Coefficient of

$x^{n}= 1+$ coefficient of

$x^{n}$ in

$(x+........ x^{n})^{2}$

$= 1+1+$ coefficient

$x^{n}$ in

$(x^{2}+...... x^{n})^{2}$

$1+1+1+$ coefficient of

$x^{n}$ in

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

$=n +$ coefficient o f

$x^{n}$ in

$(x)^{2}$

$=n$

$A$ is correct.

${c_0},{c_1},{c_2}$ denotes coefficents expansion of

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

${c_1} + {c_1}{c_2} + {c_2}{c_3} + .......{c_{n - 1}}{c_n} = \frac{{(2n)!}}{{(n + 1)!(n - 1)!}}$

0%
True
0%
False

Explanation

$C_{0}, C_{1}, C_{2}$ denotes coefficients expansion of

$(1+x)^{\text {n }}$ then

$C_{1}+C_{1} C_{2}+C_{2} C_{3}+\ldots+C_{n-1} C_{n}=\frac{2 n !}{(n-1) !(n+1) !}$

we know that

$(1+x)^{n}=C_0+C_1 x+C_{2} x^{2}+C_3 x^{3}+\cdots+C_{n-2} x^{n-2}$

$+C_{n-1} x^{n-1}+C_n x^{n}$

Also we can write it as

$(1+x)^{n}=C_n x^{n}+C_{n-1} x^{n-1}+\cdots C_2 x^{2}+C_1 x+C_0$

Multiplying these two expressions

$(1+x)^{2 n}=\left(C_{0}+C_1 x+C_{2} x^{2}+\ldots+C_{n-1}x^{n-1}+C_n x^{n})X\right.$

$\left(C_n x^{n}+C_{n-1} x^{n-1}+\ldots+C_2 x^{2}+C_{1} x+C_0)\right.$

Now equate the coefficients of

$x^{n-r}$ from both sides of equations

We get

$^{2n} C_{n-r}=C_0C_{n-r}+C_1C_{n-r-1}+C_{2}C_{n-r-2}+C_{3}C_{n-r-3}+\ldots+C_{n-r}C_n$

Put

$r=1$

$\quad C_{0} C_{1}+C_1 C_{2}+C_{2} C_{3}+\cdots+C_{n-1} C_{n}=\frac{2 n !}{(n-1) !(n+1) !}$

The sum of rational term in the expansion of

${(3^{\frac{1}{5}}+2^{\frac{1}{3}})}^{15}$ is

0%
$31$
0%
$59$
0%
$51$
0%
$61$

Explanation

Given that ,

${{\left( {{3}^{\frac{1}{5}}}+{{2}^{\frac{1}{3}}} \right)}^{15}}$

General term is

${{t}_{r+1}}{{=}^{15}}{{C}_{r}}{{\left( {{3}^{\frac{1}{5}}} \right)}^{15-r}}{{\left( {{2}^{\frac{1}{3}}} \right)}^{r}}$

${{=}^{15}}{{C}_{r}}{{.3}^{\frac{15-r}{5}}}{{.2}^{\frac{r}{3}}}$

For general term r=0,5

${{t}_{0+1}}{{=}^{15}}{{C}_{0}}{{.3}^{3}}=27$

${{t}_{15+1}}={{t}_{16}}{{=}^{15}}{{C}_{15}}{{3}^{0}}.$ ${{2}^{5}}$ =32

Hence, this is the required value = $27+32 =59$

The middle term (s) in the expansion of

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

0%
$_{ }^{ 2n+1 }{ { C }_{ n } }{ X }^{ n } and _{ }^{ 2n+1 }{ { C }_{ n+1 } }{ X }^{ n+1 }$
0%
$_{ }^{ 2n+1 }{ { C }_{ n } }{ X }^{ n+1 } and _{ }^{ 2n+1 }{ { C }_{ n+1 } }{ X }^{ n }$
0%
$^{ 2n+1 }{ { C }_{ n+1 } }{ X }^{ n }$
0%
$^{ 2n+1 }{ { C }_{ n+1 } }{ X }^{ n+1 }$

Explanation

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

Middle term

$\Rightarrow (n+1)^{th}$ term

$(n+2)^{th}$ term

$\therefore (n+2)^{th}$ term

$\Rightarrow ^{2n+1}C_{n+1}(1)^{(2n+1)-(n-1)}x^{n+1}$

$(n+2)^{th}$ term

$\Rightarrow \boxed{^{2n+1}C_{n+1}x^{n+1}}$

and

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

$= ^{2n+1}C_x (1)^{(2n+1)-(n)} x^{n}$

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

$\Rightarrow \boxed{^{2n+1}C_{n} x^{n}}$

(option A is correct)

1114538_1139611_ans_57d73fdccbd04bcb901d2219c828c27f.jpeg

$6^{th}$ term in the expansion of

$\left[\dfrac{1}{x^{8/3}}+x^{2 }\log_{10}x\right]^{8}$ is

$5600$ , then

$x$ is equal to

0%
$5$
0%
$4$
0%
$8$
0%
$none\ of\ these$

Explanation

Using binomial theorem, the 6th term of

$(a+b)^8$ is

$56a^3b^5$

Now substituting the given terms, we get

$56\cdot\dfrac{1}{(x^{8/3})^3}\cdot x^{10}\cdot(\log x)^5=5600$

$56\cdot\dfrac{1}{x^8}\cdot x^{10}\cdot(\log x)^5=5600$

$x^2(\log x)^5=100$

$x^2(\log x)^5=10^2$

$x^{\frac{2}{5}.5}(\log x)^5=10^2$

$(x^{\frac{2}{5}}\log x)^5=10^2$

$x^{\frac{2}{5}}\log x=10^{\frac{2}{5}}$

$\log x^{x^{\frac{2}{5}}}=10^{\frac{2}{5}}$

$x^{x^{\frac{2}{5}}}=10^{10^{\frac{2}{5}}}$

Comparing the both side, we get

$x=10$

$\sum _{ r=0 }^{ n } \left( \dfrac {r+2}{r+1} \right)$

$.^nCr$ is equal to :

0%
$\dfrac {2^n (n+2)-1}{(n+1)}$
0%
$\dfrac {2^n (n+1)-1}{(n+1)}$
0%
$\dfrac {2^n (n+4)-1}{(n+1)}$
0%
$\dfrac {2^n (n+3)-1}{(n+1)}$

The value of

$\displaystyle\sum^{10}_{r=0}$

$^{20}C_r$ is equal to?

0%
$\dfrac{1}{2}(2^{20}+$ $^{20}C_{10})$
0%
$\dfrac{1}{2}(2^{28}+$ $^{19}C_{10})$
0%
$20(2^{18}+$ $^{19}C_{11})$
0%
$10(2^{18}+$ $^{19}C_{11})$

Explanation

$\begin{array}{l} \sum _{ r=0 }^{ 10 }{ { \, ^{ 20 } }{ C_{ r } } } =\, { \, ^{ 20 } }{ C_{ 0 } }+\, { \, ^{ 20 } }{ C_{ 1 } }........{ \, ^{ 20 } }{ C_{ r } }={ 2^{ 20 } } \\ \left( { ^{ 20 }{ C_{ 0 } }{ +^{ 20 } }{ C_{ 1 } }{ { .... }^{ 20 } }{ C_{ 10 } } } \right) +\left( { ^{ 20 }{ C_{ 11 } }{ { ...... }^{ 20 } }{ C_{ 20 } } } \right) ={ 2^{ 20 } } \\ \left( { ^{ 20 }{ C_{ 0 } }{ { ...... }^{ 20 } }{ C_{ 10 } } } \right) +\left( { ^{ 20 }{ C_{ 9 } }{ +^{ 20 } }{ C_{ 8 } }{ { ..... }^{ 20 } }{ C_{ 0 } } } \right) ={ 2^{ 20 } } \\ =2\left( { ^{ 20 }{ C_{ 0 } }{ { ........ }^{ 20 } }{ C_{ 10 } } } \right) ={ 2^{ 20 } }{ +^{ 20 } }{ C_{ 10 } } \\ { =^{ 20 } }{ C_{ 0 } }+{ ....^{ 20 } }{ C_{ 10 } }={ 2^{ 19 } }+{ \frac { 1 }{ 2 } ^{ 20 } }{ C_{ 10 } } \end{array}$

In the binomial expansion of

$(a-b)^{n}, n\ge 5$ , the sum of the

$5^{th}$ and

$6^{th}$ terms is zero, then

$a/b$ is equals:

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

Explanation

Acc. to Question

$^{m}C_{4} a^{n-4}(-b)^{4}+ ^{n}C_{5} a^{n-5}(-b)^{5}=0$

solving we get

$a/b =\dfrac {n-4}{5}$

The first

$3$ terms in the expansion of

$(1+ax)^{n}(n\neq 0)$ are

$1, 6x$ and

$16x^{2}$ . Then the value of

$a$ and

$n$ are respectively

0%
$2$ and $9$
0%
$3$ and $2$
0%
$2/3$ and $9$
0%
$3/2$ and $6$

Explanation

$(1+ax)^n={^{n}C_o1}+{^{n}C_1}ax+{^{n}C_2}(ax)^2$

$\therefore {^{n}C_1}(ax)=6x$

$\therefore {^{n}C_2}(ax)^2=16x^2$

$\therefore \dfrac{n!}{(n-1)!}\times a=6$

$\therefore xa=6$

$\therefore \dfrac{n!a^2}{(n-2)!2!}=16$

$\therefore (n)(n-1)a^2=32$

$\therefore \dfrac{1}{(n-1)(a)}=\dfrac{3}{16}$

$\therefore \dfrac{1}{na-a}=\dfrac{3}{16}$

$\therefore 6-a=\dfrac{16}{3}$

$\therefore a=\dfrac{6-16}{3}=\dfrac{2}{3}$

$\therefore x=\dfrac{6}{2}\times 3=9$

$\therefore x=9, a=\dfrac{2}{3}$ .

1072134_1160088_ans_d0b79e08d28f45c0a8e72810b895a435.jpeg

$f ( x ) + 2 f ( 1 - x ) = x ^ { 2 } + 2 , \forall x \in R$ , then find

$f ( x )$

0%
$\dfrac{(x+3)^2}{3}$
0%
$\dfrac{(x-3)^2}{3}$
0%
$\dfrac{(x-2)^2}{4}$
0%
$\dfrac{(x-2)^2}{3}$

Explanation

Given

$f(x)+2f(1-x)=x^2+2 \cdots(1)$

Replace

$x$ with

$1-x$

$\implies f(1-x)+2f(1-(1-x))=(1-x)^2+2$

$\implies f(1-x)+2f(x)=x^2-2x+3 \cdots(2)$

$(1)+(2)$

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

$\implies f(x)+f(1-x)=\dfrac 23x^2-\dfrac 23 x+\dfrac 53 \cdots(3)$

$(2)-(1)$

$\implies f(x)-f(1-x)=1-2x \cdots(4)$

$(3)+(4)$

$\implies 2f(x)= \dfrac 23x^2-\dfrac 23 x+\dfrac 53 +1-2x$

$\implies f(x)=\dfrac 13\left(x-2\right)^2$

The number of non-zero terms in the expansion of

$(\sqrt{7}+1)^{75}-(\sqrt{7}-1)^{75}$ is

0%
$36$
0%
$37$
0%
$38$
0%
$39$

Explanation

$(\sqrt 7+1)^{75}-(\sqrt 7-1)^{75}$

$\Rightarrow \sqrt 7\left(1+\dfrac{1}{\sqrt{7}}\right)^{75}-\sqrt 7\left(1-\dfrac{1}{\sqrt{7}}\right)^{75}$

$\Rightarrow$ middle term = no of non zero terms=

$=36$

The coefficient of

${x^5}$ in the expansion of

${\left( {1 + {x^2}} \right)^5}{\left( {1 - x} \right)^4}$ is

0%
${4.^6}{C_4}$
0%
${2.^6}{C_4}$
0%
${2.^6}{C_2}$
0%
${4.^6}{C_2}$

Explanation

$\text { Coefficient of } x^{5} \text { in }\left(1+x^{2}\right)^{5}(1-x)^{4} \\$

$\left(1+x^{2}\right)^{5}={ }^{5} C_{0}\left(x^{2}\right)^{0}+{ }^{5} C_{1} x^{2}+\ldots+{ }^{5}C_{5}\left(x^{2}\right)^{5} \\$

$(1-x)^{4}={ }^{4}C_0(-1)^{0}+{ }^{4} C_{1}(-x)+\ldots+{ }^{4} C_{4} (-x)^{4} \\$

$\text { Coefficient of } x^{5} \text { will be }$

$\Rightarrow { }^{5} C_{2}\left(x^{2}\right)^{2} { }^{4} C_{1}(-x)+{ }^{5}$

$C_{1}\left(x^{2}\right){ }^{4} C_{3}(-x)^{3}$

$\Rightarrow -60 \\$

$\text { or }-2{ }^{6} C_{2}$

The number of rational terms in the expansion of

$\quad { \left( { 3 }^{ \cfrac { 1 }{ 4 } }+{ 7 }^{ \cfrac { 1 }{ 6 } } \right) }^{ 144 }$ is

0%
$33$
0%
$23$
0%
$12$
0%
$13$

Explanation

$(3^{1/4}+7^{1/6})^{144}$

general term

$={^{144}C_r}3^{\dfrac{1}{4}(144-r)}7^{1/6(r)}$

$={^{144}C_r}3^{36-\dfrac{r}{4}}7^{\dfrac{r}{6}}$

For rational terms

$r=$ LCM

$(4, 6)=12$

$r=0, 12, 24, 36,.....144$

$\rightarrow$

$144=0+(n-1)12$

$n-1=\dfrac{144}{2}=12$

$n=13$

$\therefore$

$13$ rational term.

1054665_1158993_ans_5773a3851c8948c8bbbf2dc300a2d8b5.jpg

In the expansion of

$(y^{1/5}+x^{1/10})^{55}$ , the number of terms free of a radical sign is

0%
$5$
0%
$6$
0%
$50$
0%
$56$

Explanation

$\left(y^{1 / 5}+x^{1 / 10}\right)^{55}$

$T_{r+1}=55C_r\left(y^{1 / 5}\right)^{r}\left(x^{1 / 10}\right)^{55-r}$

This will be independent of radicals after making powers of

$x, y$ integers

$\frac{r}{5}$ and

$\frac{r}{10}$ should be integers

$\therefore r=0,10,20,30,40,50$

6 terms will be frel of radicals.

For

$r=0,1,2,,....10$ let

${A}_{r},{B}_{r}$ and

${C}_{r}$ denote respectively the coefficient of

${x}^{r}$ in the expansions of

${(1+x)}^{10},{(1+x)}^{20}$ and

${(1+x)}^{30}$ . Then

$\sum _{ r=1 }^{ 10 }{ { A }_{ r }\left( { B }_{ 10 }{ B }_{ r }-{ C }_{ 10 }{ A }_{ r } \right) }$ is equal to

0%
${B}_{10}-{C}_{10}$
0%
${A}_{10}({B}_{10}^{2}-{C}_{10}{A}_{10})$
0%
$0$
0%
${C}_{10}-{B}_{10}$

Explanation

$A_{r}=$ Coefficient of

$x^{r}$ in

$(1+r)^{10}=10\ C_{r}$

$B_{r}=$ Coefficient of

$x^{r}$ in

$(1+r)^{20}=20\ C_{r}$

$C_{r}=$ Coefficient of

$x^{r}$ in

$(1+r)^{30}=30\ C_{r}$

$\displaystyle\sum_{r=1}^{10}Ar (B_{10}B_{r}-C_{10}A_{r})=\sum_{r=1}^{10}A_{r}B_{10}B_{r}-\sum_{r=1}^{10}A_{r}C_{10}A_{r}$

$=\displaystyle\sum_{r=1}^{10} \_{}^{10}C_{r} \_{}^{20}C_{10} \_{}^{20}C_{r}-\sum_{r=1}^{10} \_{}^{10}C_{r} \_{}^{30}C_{10} \_{}^{10}C_{r}$

$= \_{}^{20}C_{10}\displaystyle\sum_{r=1}^{10} \_{}^{10}C_{10-r} \_{}^{20}C_{r}- \_{}^{30}C_{10}\sum_{r=1}^{10} \_{}^{10}C_{10-r} \_{}^{10}C_{r}$

$=_{}^{20}C_{10}\left(_{}^{30}C_{10}-1\right)-_{}^{30}C_{10}\left(_{}^{20}C_{10}-1\right)$

$=_{}^{30}C_{10}-_{}^{20}C_{10}=C_{10}-B_{10}$

$C_{10}-B_{10}$

The coefficient of the middle term in the expansion of

$\left(1+x\right)^{2n}$ is equal to the sum of the coefficient of middle terms in the expansion of

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

the statement is true and false

0%
True
0%
False

The numerical value of middle terms in

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

0%
${ _{ }^{ 2n }{ C } }_{ n }$
0%
${ _{ }^{ n }{ C } }_{ n }$
0%
$\left( { _{ }^{ 2n }{ C } }_{ n } \right)$
0%
$\left( { _{ }^{ n }{ C } }_{ n } \right)$

Explanation

$(1-\dfrac{1}{x})^{n}(1-x)^{n}$

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

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

middle term =

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

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

middle term =

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

The value of

$\left( { _{ }^{ 7 }{ C } }_{ 0 }+{ _{ }^{ 7 }{ C } }_{ 1 } \right) +\left( { _{ }^{ 7 }{ C } }_{ 1 }+{ _{ }^{ 7 }{ C } }_{ 2 } \right) +.....\left( { _{ }^{ 7 }{ C } }_{ 6 }+{ _{ }^{ 7 }{ C } }_{ 7 } \right)$ is

0%
${2}^{7}-1$
0%
${2}^{8}-2$
0%
${2}^{8}-1$
0%
${2}^{8}$

Explanation

$(^{7}C_{0}+^{7}C_{1}+...^{7}C_{6})+(^{7}C_{1}+^{7}C_{2}+...^{7}C_{7})$

$(2^{7}-^{7}C_{7})+(2^{7}-^{7}C_{0})$

$(2^{7}-1)+(2^{7}-1)$

$2.2^{7}-2$

$2^{8}-2$

1082240_1172873_ans_3cb46481271d4e9ba983d1d2cb6f5908.jpg

The number of rational terms in the expansion of

${ \left( \sqrt [ 4 ]{ 5 } +\sqrt [ 5 ]{ 4 } \right) }^{ 100 }$ is

0%
$50$
0%
$5$
0%
$6$
0%
$51$

The

$r$ th term of series

$2\dfrac{1}{2} + 1\dfrac{7}{{13}} + 1\dfrac{1}{9} + \dfrac{{20}}{{23}} + .....$ is

0%
$\dfrac{{20}}{{5r + 3}}$
0%
$\dfrac{{20}}{{5r - 3}}$
0%
$20\left( {5r + 3} \right)$
0%
$\dfrac{{20}}{{5{r^2} + 3}}$

$^{m}C_r.^{n}C_0+{}^{m}C_{r-1}.{}^{n}C_{1}+{}^{m}C_{r-2}.^{n}C_2+..........+^{m}C_0.^{n}C_r={}^{m+n}Cr$

0%
True
0%
False

$r$ th term is middle term in

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

$(r+3)$ th term is:

0%
$\cfrac { { _{ }^{ 20 }{ C } }_{ 7 }x }{ { 2 }^{ 13 } }$
0%
$-\left( \cfrac { { _{ }^{ 20 }{ C } }_{ 5 }x }{ { 4 }^{ 13 } } \right)$
0%
$-\left( \cfrac { { _{ }^{ 20 }{ C } }_{ 7 }x }{ { 2 }^{ 13 } } \right)$
0%
$-\left( \cfrac { { _{ }^{ 20 }{ C } }_{ 14 }x }{ { 4 }^{ 13 } } \right)$

The middle term in the expansion of

$(1-3x+3x^{2}-x^{3})^{6}$ is

0%
$^{18}C_{10}\ {x}^{10}$
0%
$^{18}C_{9}( {-x})^{9}$
0%
$^{18}C_{9}\ {x}^{9}$
0%
$None\ of\ these$

The value of

$\displaystyle\frac { { C }_{ 0 } }{ 1.3 } -\frac { { C }_{ 1 } }{ 2.3 } +\frac { { C }_{ 2 } }{ 3.3 } -\frac { { C }_{ 3 } }{ 4.3 } +.........+{ \left( -1 \right) }^{ n }\frac { { C }_{ n } }{ (n+1).3 }$ is :

0%
$\displaystyle\frac { 3 }{ n+1 }$
0%
$\displaystyle\frac { n+1 }{ 3 }$
0%
$\displaystyle\frac { 1 }{ 3n+3 }$
0%
none of these

The sum of the series

$^{20}{C_0}{ - ^{20}}{C_1}{ + ^{20}}{C_2}{ - ^{20}}{C_3}{ + _{......}}{ - _{......}}{ + ^{20}}{C_{10}}\,is -$

0%
$\frac{1}{2}{20_{{C_{10}}}}$
0%
$0$
0%
${ - ^{20}}{C_{10}}$
0%
$^{20}{C_{10}}$

Explanation

We know that,

$(1+x)^{20}=^{20}{C_0}{ + ^{20}}{C_1}x{ + ^{20}}{C_2}x^2{ + ^{20}}{C_3}x^3{ + _{......}}{ _{......}}{ + ^{20}}{C_{10}}x^{10}+_{.............} +^{20}C_{20}x^{20}$

Putting

$x=-1$ we get

$0=^{20}{C_0}{ - ^{20}}{C_1}{ + ^{20}}{C_2}{ - ^{20}}{C_3}{ + _{......}}{ _{......}}{ + ^{20}}{C_{10}}+_{.............} +^{20}C_{20}$

But as we know , that

$^{20}C_0=^{20}C_{20}, ^{20}C_1=^{20}C_{19}$ and so on , we can write above equation as

$0=2(^{20}{C_0}{ - ^{20}}{C_1}{ + ^{20}}{C_2}{ - ^{20}}{C_3}+{..........}-^{20}C_9)+^{20}C_{10}\\ -\dfrac{^{20}C_{10}}{2}=(^{20}{C_0}{ - ^{20}}{C_1}{ + ^{20}}{C_2}{ - ^{20}}{C_3}+{..........}-^{20}C_9)$

Hence,

$^{20}{C_0}{ - ^{20}}{C_1}{ + ^{20}}{C_2}{ - ^{20}}{C_3}{ + _{......}}{ - _{......}}{ + ^{20}}{C_{10}}=-\dfrac{^{20}C_{10}}{2}{ + ^{20}}{C_{10}}=\dfrac{^{20}C_{10}}{2}$

If the number of terms is the expansion

${\left( {1 - \frac{2}{x} + \frac{4}{{{x^2}}}} \right)^n},x \ne 0,$ is

$28$ , then the sum of coefficients of all the terms in this expansion, is :

0%
2187
0%
243
0%
729
0%
64

The sum of coefficients of integral powers of

$x$ in the binomial expansion of

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

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

Explanation

${ \left( 1+2\sqrt { x } \right) }^{ 50 }={ _{ }^{ 50 }{ C } }_{ 0 }{ \left( 1 \right) }^{ }+{ _{ }^{ 50 }{ C } }_{ 1 }{ \left( 2\sqrt { x } \right) }^{ }+{ _{ }^{ 50 }{ C } }_{ 2 }{ \left( 2\sqrt { x } \right) }^{ 2 }+.....$

$I={ _{ }^{ 50 }{ C } }_{ 0 }+{ _{ }^{ 50 }{ C } }_{ 2 }\left( { 2 }^{ 2 } \right) +{ _{ }^{ 50 }{ C } }_{ 4 }\left( { 2 }^{ 4 } \right) +...+{ _{ }^{ 50 }{ C } }_{ n }\left( { 2 }^{ 2n } \right)$

${ \left( 1+2 \right) }^{ 50 }={ _{ }^{ 50 }{ C } }_{ 0 }+{ _{ }^{ 50 }{ C } }_{ 1 }(2)+{ _{ }^{ 50 }{ C } }_{ 2 }\left( { 2 }^{ 2 } \right) +....+{ _{ }^{ 50 }{ C } }_{ n }\left( { 2 }^{ n } \right)$

${ 3 }^{ 50 }=I+2\left[ { _{ }^{ 50 }{ C } }_{ 0 }+{ _{ }^{ 50 }{ C } }_{ 3 }\left( { 2 }^{ 2 } \right) +{ _{ }^{ 50 }{ C } }_{ (2n+1) }\left( { 2 }^{ 2n } \right) \right]$

${ 3 }^{ 50 }=I+\left[ { _{ }^{ 50 }{ C } }_{ 0 }-{ _{ }^{ 50 }{ C } }_{ 0 }+{ _{ }^{ 50 }{ C } }_{ 1 }(2)+{ _{ }^{ 50 }{ C } }_{ 3 }\left( { 2 }^{ 3 } \right) +....+{ _{ }^{ 50 }{ C } }_{ { (2n+1) }^{ 2 } }\left( { 2 }^{ 2n+1 } \right) \right]$

${ 3 }^{ 50 }==I+I-1$

$\left( { 3 }^{ 50 }+1 \right) =2I\Rightarrow I=\cfrac { 1 }{ 2 } \left( { 3 }^{ 50 }+1 \right)$

$A$ and

$B$ are coefficients of

$x ^ { n }$ in the expansion of

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

$( 1 + x ) ^ { 2 n - 1 }$ respectively, then

0%
$A = B$
0%
$A = 2 B$
0%
$2 A = B$
0%
$A + B = 0$

$(1+x)^n=C_0+C_1x+C_2x^2+......+C_nx^n,$ then

$C_0+5C_1+9C_2+.....+(4n+1)C_n$ is equal to

0%
$n.2^n$
0%
$(n+1)2^n$
0%
$(2n+1)2^n$
0%
$(4n+1)2^n$

Explanation

$\frac{\left[C_{0}+C_{1}+C_{2}+\ldots+C_{n}\right]+\varphi_{1}+8 C_{2}+\ldots+4 n(n]}{2^{n}+4\left[C_{1}+2 C_{2}+3 C_{3}+\ldots+n C_{n}\right]}$

$(1+x)^{n}=C_{0}+C_{1} x+\ldots+\ln x^{n}$

differentiate on both sides

$C_{1}+2 C_{2}+3 C_{3}+\ldots+nC_{n}=n(1+x)^{n-1}$

Put

$x=1 \Rightarrow n(2)^{n-1}$

Now puting value in equation

$2^{n}+4\left[n 2^{n-1}\right] \Rightarrow 2^{n}(2 n+1)$

If the rth term in the expansion of

${\left( {{x \over 3} - {2 \over {{x^2}}}} \right)^{10}}$ contains

${x^4}$ then r is equal to

0%
2
0%
3
0%
4
0%
5

If the sum of the binomial coefficients in the expansion of

$\left(x^{2}+\dfrac{2}{x^{3}}\right)^{n}$ is

$243$ , the term independent of

$x$ is equal to

0%
$40$
0%
$30$
0%
$20$
0%
$10$

$x^{4}$ occurs in the

$rth$ term in the expansion of

$\left(x^{4}+\dfrac{1}{x^{3}}\right)^{15}$ , then

$r=$

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

Explanation

$\left(x^4+\dfrac{1}{x^3}\right)^{15}=^{15}C_0 (x^4)^0\times \left(\dfrac{1}{x^3}\right)^{15}+........+^{15}C_r (x^4)^r\left(\dfrac{1}{x^3}\right)^{15-r}+.......$

Thus

$T_{r+1}=^{15}C_r x^{4r}\left(\dfrac{1}{x}\right)^{45-3r}$

$T_{r+1}=^{15}C_4 x^{7r-45}$

to get

$x^4\Rightarrow 7r-45=4$

$7r=49$

$r=7$

Thus

$T_8=^{15}C_4 x^4$

$(1-x-x^2)^{20}$ =

$\sum _{ r=0 }^{ 40 }{ a_4,x^x },$ then

$a_1+3a_3+5a_5+........+39a_{39}=$

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

whether the sum of the coefficients in the expansion of

${\left(1+x-3{x}^{2}\right)}^{2163}$ is

$-6$

0%
True
0%
False

If the constant term in the expansion of

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

$15$ then the value of

$n$ is

0%
$6$
0%
$9$
0%
$12$
0%
$15$

In the expansion of

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

$x^{4}$ is not divisible by

0%
$12$
0%
$7$
0%
$11$
0%
$5$

Coefficient of

$\alpha$ in the expansion of

$(\alpha +p)^{m-1}+(\alpha +p)^{m-2}(\alpha +q)^{m-3}(\alpha +q)^{2}+....(\alpha +q)^{m-1}$ where

$\alpha \neq -q$ and

$p\neq q$ is:

0%
$\frac{^{m}C_{1}(p^{1}-q^{1})}{p-q}$
0%
$\frac{^{m}C_{1}(p^{m-1}-q^{m-1})}{p-q}$
0%
$\frac{^{m}C_{1}(p^{1}+q^{1})}{p-q}$
0%
$\frac{^{m}C_{1}(p^{m-1}+q^{m-1})}{p-q}$

If the

$r^{th}$ and the

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

$(p+q)^{n}$ are equal, then

$\dfrac{(n+1)q}{r(p+q)}$ is

0%
$1/2$
0%
$1/4$
0%
$1$
0%
$0$

Number of distinct terms in the expansion of

$(x+y-z)^{16}$ is

0%
$816$
0%
$152$
0%
$153$
0%
$136$

Find the coefficient of

$x^{11}$ in the expansion of

$\left(x^{3}-\dfrac{2}{x^{2}}\right)^{12}$

0%
$-25344$
0%
$-25250$
0%
$-25000$
0%
$-25750$

The number of terms in the expansion of

$(1+x)^{101}(1+x^{2}-x)^{100}$ in power of

$x$ is:

0%
$302$
0%
$301$
0%
$202$
0%
$101$

$(1+x+x^2)^8=a_0+a_1x+.....a_{16}x^{16}$ then

$a_1-a_3+a_5-a_7+....... -a_{15}$ =

0%
$1$
0%
$2$
0%
$3$
0%
$0$

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 8 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 8 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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