Binomial Theorem - For Positive Integral Index (Lecture-02)

Summation of Series (involving binomial coefficients)

1. Bino-geometric series

${ }^{n} \mathrm{C} _{0}+{ }^{n} \mathrm{C} _{1} \mathrm{x}+{ }^{\mathrm{n}} \mathrm{C} _{2} \mathrm{x}^{2}$.

$+{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}} \mathrm{x}^{\mathrm{n}}=(1+\mathrm{x})^{\mathrm{n}}$

2. Bino-arithmetic series

$a^{n} C _{o}+(a+d)^{n} C _{1}+(a+2 d)^{n} C _{2}+\ldots \ldots \ldots \ldots \ldots \ldots . .+(a+n d)^{n} C _{n}$

This series is the sum of the products of corresponding terms of

${ }^{\mathrm{n}} \mathrm{C} _{0},{ }^{\mathrm{n}} \mathrm{C} _{1},{ }^{\mathrm{n}} \mathrm{C} _{2}, \ldots \ldots \ldots \ldots . .{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}}$ (binomial coefficients) and $\mathrm{a}, \mathrm{a}+\mathrm{d}, \mathrm{a}+2 \mathrm{~d}$ ……………a+nd

(arithmetic progression)

Such series can be solved either by

(i) eliminating $\mathrm{r}$ in the multiplier of binomial coefficient from the $(\mathrm{r}+1)^{\text {th }}$ terms of the series (i.e. using $\mathrm{r}^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}=\mathrm{n}^{\mathrm{n}-1} \mathrm{C} _{\mathrm{r}-1}$ )

or

(ii) Differentiating the expansion of $\mathrm{x}^{\mathrm{a}}\left(1+\mathrm{x}^{\mathrm{d}}\right)^{\mathrm{n}}$ or (If product of two or more numericals occur, then differentiate again and again till we get the desired result)

3. Bino-harmonic series

$\dfrac{{ }^{n} C _{0}}{a}+\dfrac{{ }^{n} C _{1}}{a+d}+\dfrac{{ }^{n} C _{2}}{a+2 d}+\ldots \ldots . .+\dfrac{{ }^{n} C _{n}}{a+n d}$

This series is the sum of the products of corresponding terms of

${ }^{\mathrm{n}} \mathrm{C} _{0},{ }^{\mathrm{n}} \mathrm{C} _{1},{ }^{\mathrm{n}} \mathrm{C} _{2}, \ldots \ldots \ldots \ldots . .{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}}$ (binomial coefficients) and

$\dfrac{1}{a}, \dfrac{1}{a+d}, \dfrac{1}{a+2 d}, \ldots \ldots \ldots \ldots \ldots . . . \dfrac{1}{a+n d}$ (harmonic progression)

Such seris can be solved either by

(i) eliminating $\mathrm{r}$ in the multiplier of binomial coefficient from the $(\mathrm{r}+1)^{\text {th }}$ term of the series

(ie using $\dfrac{1}{\mathrm{r}+1}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}=\dfrac{1}{\mathrm{n}+1}{ }^{\mathrm{n}+1} \mathrm{C} _{\mathrm{r}+1}$ ) or

(iii) integrating suitable expansion

Note

(i) If the sum contains $\mathrm{C} _{0}, \mathrm{C} _{1}, \mathrm{C} _{2} \ldots \ldots \ldots \ldots \mathrm{C} _{\mathrm{n}}$ are all positive signs, integrate between limits 0 to 1

(ii) If the sum contains alternate signs (i.e. + & ) then integrate between limits -1 to 0

(iii) If the sum contains odd coefficients (i.e. $\mathrm{C} _{0}, \mathrm{C} _{2}, \mathrm{C} _{4}, \ldots \ldots .$. ) then integrate between -1 to +1 .

(iv) If the sum contains even coefficient (i.e. $\mathrm{C} _{1}, \mathrm{C} _{3}, \mathrm{C} _{5}, \ldots \ldots .$. ) then find the difference between (i) & (iii) and then divide by 2

(v) If in denominator of binomial coefficient is product of two numericals then integrate two times first time take limits between 0 to $x$ and second time take suitable limits

4. Bino-binomial series.

${ }^{\mathrm{n}} \mathrm{C} _{0}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C} _{1}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}+1}+{ }^{\mathrm{n}} \mathrm{C} _{2}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}+2}+\ldots \ldots \ldots .+{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}-\mathrm{r}}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}} \quad$ or

${ }^{\mathrm{m}} \mathrm{C} _{0}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}+{ }^{\mathrm{m}} \mathrm{C} _{1}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}-1}+{ }^{\mathrm{m}} \mathrm{C} _{2}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}-2}+\ldots \ldots \ldots . .+{ }^{\mathrm{m}} \mathrm{C} _{\mathrm{r}}{ }^{\mathrm{n}} \mathrm{C} _{0}$

Such series can be solved by multiplying two expansions, one involving the first factors as coefficient and the other involving the second factors as coefficients and finally equating coefficients of a suitable power of $x$ on both sides.

Divisibility Problems

To show that an expression is divisible by an integer

  • write $a^{p n+r}=a^{p n} a^{r}=\left(a^{p}\right)^{n} a^{r}$ if a, $p, n, r \in N$
  • to show the expression is divisible by $\mathrm{k}$, express $\mathrm{a}^{\mathrm{p}}=1+\left(\mathrm{a}^{\mathrm{p}}-1\right)$; if some power of $\left(\mathrm{a}^{\mathrm{p}}-1\right)$ has $\mathrm{k}$ as a factor $a^{p}=2+\left(a^{p}-2\right)$; if some power of $\left(a^{p}-2\right)$ has $k$ as a factor

$a^{p}=q+\left(a^{p}-q\right) \text { if some power of }\left(a^{p}-q\right) \text { has } k \text { as a factor }$

Binomial coefficients

1. $\mathrm{C} _0+\mathrm{C} _1+\mathrm{C} _2+\mathrm{C} _3+\ldots \ldots \ldots \ldots 2^{\mathrm{n}}$

2. $\mathrm{C} _0-\mathrm{C} _1+\mathrm{C} _2-\mathrm{C} _3+\ldots \ldots \ldots \ldots=0$

3. $\mathrm{C} _0-\mathrm{C} _1+\mathrm{C} _2-\mathrm{C} _3+\ldots \ldots \ldots+\mathrm{C} _{\mathrm{r}}(-1)^{\mathrm{r}}={ }^{\mathrm{n}-1} \mathrm{C} _{\mathrm{r}}(-1)^{\mathrm{r}} ; \mathrm{r}<\mathrm{n}$

4. $\mathrm{C} _0+\mathrm{C} _2+\mathrm{C} _4+\mathrm{C}_6+\ldots \ldots \ldots \ldots=2^{\mathrm{n}-1}$

5. $\mathrm{C} _1+\mathrm{C} _3+\mathrm{C} _5+\mathrm{C}_7+\ldots \ldots \ldots \ldots=2^{\mathrm{n}-1}$

6. $\mathrm{C} _0-\mathrm{C} _2+\mathrm{C} _4-\mathrm{C}_6+\ldots \ldots \ldots=(\sqrt{2})^{\mathrm{n}} \cos \dfrac{\mathrm{n} \pi}{4}$

7. $\mathrm{C} _1-\mathrm{C} _3+\mathrm{C} _5-\mathrm{C}_7+\ldots \ldots=(\sqrt{2})^{\mathrm{n}} \sin \dfrac{\mathrm{n} \pi}{4}$

8. $\mathrm{C} _0+\mathrm{C} _4+\mathrm{C} _8+\mathrm{C} _{12}+\ldots \ldots=\dfrac{1}{2}\left(2^{\mathrm{n}-1}+\sqrt{2}^{\mathrm{n}} \cos \dfrac{\mathrm{n} \pi}{4}\right)$

9. $\mathrm{C} _1+\mathrm{C} _5+\mathrm{C} _9+\mathrm{C} _{13}+\ldots \ldots=\dfrac{1}{2}\left(2^{\mathrm{n}-1}+\sqrt{2}^{\mathrm{n}} \sin \dfrac{\mathrm{n} \pi}{4}\right)$

10. $\mathrm{C} _0+\mathrm{C} _3+\mathrm{C} _6+\mathrm{C} _9+\ldots \ldots=\dfrac{1}{3}\left(2^{\mathrm{n}}+2 \cos \dfrac{\mathrm{n} \pi}{3}\right)$

11. $\mathrm{C} _1+2 \mathrm{C} _2+\ldots \ldots \ldots \ldots=\sum \mathrm{rC} _{\mathrm{r}}=\mathrm{n} \cdot 2^{\mathrm{n}-1}$

12. $\mathrm{C} _1-2 \mathrm{C} _2+3 \mathrm{C} _3 \ldots \ldots \ldots \ldots \ldots=\sum(-1)^{\mathrm{r}-1} \mathrm{r} \mathrm{C} _{\mathrm{r}}=0$

13. $1^2 \mathrm{C}_1+2^2 \mathrm{C}_2+\ldots \ldots \ldots \ldots \ldots \ldots=n(\mathrm{n}+1) 2^{2 \mathrm{n}-2}$

14. $1^2 \mathrm{C}_1-2^2 \mathrm{C}_2 \ldots \ldots \ldots \ldots=0$

15. $\mathrm{C} _0{ }^2+\mathrm{C} _1{ }^2+{ }^2 \mathrm{C} _2{ }^2+\ldots \ldots \ldots \ldots . .={ }^{2 \mathrm{n}} \mathrm{C} _{\mathrm{n}}$

16. $\mathrm{C} _0{ }^2-\mathrm{C} _1{ }^2+\mathrm{C} _2{ }^2-\mathrm{C} _3{ }^2 \ldots \ldots \ldots \ldots \ldots=\left\{\begin{array}{l}0 \text { if } \mathrm{n} \text { is odd } \\ (-1)^{\mathrm{n} / 2}{ }^n \mathrm{C} _{\mathrm{n} / 2} \text { if } \mathrm{n} \text { is even }\end{array}\right.$

17. $\sum_{0 \leq i<j \leq n} C_i C_j=2^{2 n-1}-{ }^{2 n-1} C_n$

18. $\sum_{0 \leq i<j \leq n}\left(C_i-C_j\right)^2=(n+1)^{2 n} C_n-2^{2 n}$

Note: Consider the equation $\mathrm{x} _{1}+\mathrm{x} _{2}+\ldots \ldots \ldots+\mathrm{x} _{\mathrm{r}}=\mathrm{n}, . \mathrm{n} \in \mathrm{N}$.

Number of positive integral solutions $={ }^{n-1} \mathrm{C} _{\mathrm{r}-1}$

Number of non negative integral solutions $={ }^{n+r-1} C _{r-1}$

Solved Examples

1. The value of $\sum\limits _{\mathrm{r}=1}^{\mathrm{n}}{ }^{2 \mathrm{n}} \mathrm{C} _{\mathrm{r}} \mathrm{r}$ is

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

(b) $2^{2 \mathrm{n}-1}$

(c) $2^{\mathrm{n}-1}+1$

(d) None of these

Show Answer

Solution:

$\begin{aligned} & \sum\limits _{r=1}^{n} r \dfrac{2 n}{r}{ }^{2 n-1} C _{r-1} \\ = & 2 n \sum\limits _{r=1}^{n}{ }^{2 n-1} C _{r-1} \\ = & 2 n\left({ }^{2 n-1} C _{0}+{ }^{2 n-1} C _{1}+{ }^{2 n-1} C _{2}+\ldots . .{ }^{2 n-1} C _{n-1}\right) \\ = & 2 n \cdot \dfrac{2^{2 n-1}}{2}=n \cdot 2^{2 n-1}\left[\because 2^{2 n-1}=C _{0}+C _{1}+\ldots \ldots . .+C _{2 n-1} \Rightarrow 2^{2 n-1}=2\left(C _{0}+C _{1}+C _{2}+\ldots \ldots . C _{n-1}\right)\right] \end{aligned}$

Answer (a)

2. The coefficient of $x^{5}$ in the expansion of $(1+x)^{21}+(1+x)^{22}+\ldots \ldots \ldots+(1+x)^{30}$ is

(a) ${ }^{31} \mathrm{C} _{5}{ }^{21} \mathrm{C} _{5}$

(b) ${ }^{31} \mathrm{C} _{6}{ }^{-21} \mathrm{C} _{6}$

(c) ${ }^{30} \mathrm{C} _{6}-{ }^{20} \mathrm{C} _{6}$

(d) None of these

Show Answer

Solution:

Co-efficient of $\mathrm{x}^{5}$ in $(1+\mathrm{x})^{21}+(1+\mathrm{x})^{22+}$ $+(1+\mathrm{x})^{30}$

$=$ Co-efficient of $x^{5}$ in $\dfrac{(1+x)^{21}\left{(1+x)^{10}-1\right}}{(1+x)-1}$

$\Rightarrow$ coefficient of $\mathrm{x}^{6}$ in $(1+\mathrm{x})^{31}-(1+\mathrm{x})^{21}$ is ${ }^{31} \mathrm{C} _{6}-{ }^{21} \mathrm{C} _{6}$

Answer (b)

3. The number of distinct terms in the expansion of $(x+y-z)^{16}$ is

(a) 136

(b) 153

(c) 16

(d) 17

Show Answer

Solution:

Apply ${ }^{n+r-1} \mathrm{C} _{\mathrm{r}-1}$ to get number of terms ${ }^{16+3-1} \mathrm{C} _{3-1}={ }^{18} \mathrm{C} _{2}=153$

Answer (b)

4. If I is the integral part of $(2+\sqrt{3})^{\mathrm{n}}$ and $\mathrm{f}$ is the fractional part. Then $(\mathrm{I}+\mathrm{f})(1-\mathrm{f})$ is equal to

(a) 0

(b) 1

(c) $\mathrm{n}$

(d) None of these

Show Answer

Solution:

Let $(2+\sqrt{3})^{\mathrm{n}}=\mathrm{I}+\mathrm{f}$ ………………(1) $\quad \quad \quad$ $(0 \leq \mathrm{f}<1)$

and $(2-\sqrt{3})^{\mathrm{n}}=\mathrm{F}$ ………………(2) $\hspace {1.5 cm}$ $(0<\mathrm{F}<1)$

(1) + (2) gives $\hspace {4 cm}$ (adding $0<\mathrm{f}+\mathrm{F}<2$ )

$2\left({ }^{n} \mathrm{C} _{0} \cdot 2^{\mathrm{n}}+{ }^{\mathrm{n}} \mathrm{C} _{2} \cdot 2^{\mathrm{n}-2}(\sqrt{3})^{2}+\ldots \ldots \ldots\right)=\mathrm{I}+\mathrm{f}+\mathrm{F}$

$\Rightarrow \mathrm{I}+\mathrm{f}+\mathrm{F}$ is an even integer

$\Rightarrow \mathrm{I}+\mathrm{F}$ is an integer

$\Rightarrow \mathrm{f}+\mathrm{F}=1$

$\mathrm{F}=1-\mathrm{f}$ $\hspace {4.5 cm}$ $(\because 0<\mathrm{f}+\mathrm{F}<2)$

$\therefore(\mathrm{I}+\mathrm{F})(1-\mathrm{f})=(2+\sqrt{3})^{\mathrm{n}}(2-\sqrt{3})^{\mathrm{n}}=(4-3)^{\mathrm{n}}=1$

Answer (b)

5. If the middle term of $(1+x)^{2 n}(n \in N)$ is the greatest term of the expansion, then the interval in which $\mathrm{x}$ lies is

(a) $\left[\dfrac{\mathrm{n}+1}{\mathrm{n}}, \dfrac{\mathrm{n}+2}{\mathrm{n}}\right]$

(b) $\left[\dfrac{\mathrm{n}-1}{\mathrm{n}}, \dfrac{\mathrm{n}+1}{\mathrm{n}}\right]$

(c) $\left[\dfrac{\mathrm{n}}{\mathrm{n}+1}, \dfrac{\mathrm{n}+1}{\mathrm{n}}\right]$

(d) None of these

Show Answer

Solution:

$\begin{aligned} & T _{n} \leq T _{n+1} \\ & \Rightarrow \quad{ }^{2 n} C _{n-1}^{n} x^{n-1} \leq \leq^{2 n} C _{n} \cdot x^{n} \\ & & \\ & \Rightarrow \quad \dfrac{{ }^{2 n} C _{n-1}}{{ }^{2 n} C _{n}} \leq x \\ & & \\ & T _{n+1} \geq T _{n+2} \\ & { }^{2 n} C _{n} \cdot x^{n} \geq{ }^{n+2} C _{n+1} x^{n+1} \\ & \dfrac{{ }^{2 n} C _{n}}{{ }^{2 n} C _{n+1}} \geq x \\ & x \geq \dfrac{n}{2 n-n+1} \\ & & \\ & x \leq \dfrac{n+1}{2 n-n} \\ & \Rightarrow \quad \mathrm{x} \in\left[\dfrac{\mathrm{n}}{\mathrm{n}+1}, \dfrac{\mathrm{n}+1}{\mathrm{n}}\right] \end{aligned}$

Answer (c)

6. If $\mathrm{C} _0, \mathrm{C} _1, \mathrm{C} _2, \ldots \ldots \ldots \ldots \ldots \ldots \mathrm{C} _{\mathrm{n}}$ are the binomial coefficients in expansion of $(1+\mathrm{x})^{\mathrm{n}}, \mathrm{n}$ being even, then $\mathrm{C} _0+\left(\mathrm{C} _0+\mathrm{C} _1\right)+\left(\mathrm{C} _0+\mathrm{C} _1+\mathrm{C} _2\right)+\ldots \ldots \ldots+\left(\mathrm{C} _0+\mathrm{C}_1+\ldots \ldots+\mathrm{C} _{0-1}\right)$ is equal to

(a) $\mathrm{n} \cdot 2^{\mathrm{n}}$

(b) $\text { n. } 2^{\mathrm{n}-1}$

(c) $\mathrm{n} .2^{\mathrm{n}-2}$

(d) $n .2^{n-3}$

Show Answer

Solution:

$\begin{aligned} & \therefore \mathrm{C} _{0}+\left(\mathrm{C} _{0}+\mathrm{C} _{1}\right)+\ldots \ldots \ldots+\left(\mathrm{C} _{0}+\mathrm{C} _{1}+\ldots \ldots . \mathrm{C} _{\mathrm{n}-2}\right)+\left(\mathrm{C} _{0}+\mathrm{C} _{1}+\ldots \ldots \ldots \mathrm{C} _{\mathrm{n}-1}\right) \\ & =\left(\mathrm{C} _{\mathrm{n}}\right)+\left(\mathrm{C} _{\mathrm{n}}+\mathrm{C} _{\mathrm{n}-1}\right)+\ldots \ldots .+\left(\mathrm{C} _{0}+\mathrm{C} _{1}+\ldots \ldots . .+\mathrm{C} _{\mathrm{n}-2}\right)+\left(\mathrm{C} _{0}+\mathrm{C} _{1}+\ldots . .+\mathrm{C} _{\mathrm{n}-1}\right) \\ & =2^{\mathrm{n}}+2^{\mathrm{n}}+2^{\mathrm{n}}+\ldots \dfrac{\mathrm{n}}{2} \text { times (Adding the terms equidistant from the begining and the end) } \\ & =\dfrac{\mathrm{n}}{2} \cdot 2^{\mathrm{n}}=\mathrm{n} \cdot 2^{\mathrm{n}-1} \end{aligned}$

Answer (b)

7. The number of terms in the expansion of $\left(\mathrm{x}^{3}+\dfrac{1}{\mathrm{x}^{3}}+1\right)^{100}$ is

(a) 201

(b) 200

(c) 300

(d) 100 $\mathrm{C} _{3}$

Show Answer

Solution:

$\left(\mathrm{x}^{3}+\dfrac{1}{\mathrm{x}^{3}}+1\right)^{100}=\mathrm{C} _{0}+\mathrm{C} _{1}\left(\mathrm{x}^{3}+\dfrac{1}{\mathrm{x}^{3}}\right)+\mathrm{C} _{2}\left(\mathrm{x}^{3}+\dfrac{1}{\mathrm{x}^{3}}\right)^{2}+\ldots \ldots \ldots . .+\mathrm{C} _{100}\left(\mathrm{x}^{3}+\dfrac{1}{\mathrm{x}^{3}}\right)^{100}$

gives terms of $\mathrm{x}^{3}, \mathrm{x}^{6}, \ldots \ldots \ldots \ldots \ldots \mathrm{x}^{300}, \dfrac{1}{\mathrm{x}^{3}}, \dfrac{1}{\mathrm{x}^{6}}, \ldots \ldots \ldots \dfrac{1}{\mathrm{x}^{300}}$ and a constant term

$\therefore 201$ terms

Answer (a)

Exercise

1. If $\mathrm{C} _{\mathrm{r}}$ stands for ${ }^{n} \mathrm{C} _{\mathrm{r}}$, then the sum of the series $2 \dfrac{\left(\dfrac{\mathrm{n}}{2}\right) !\left(\dfrac{\mathrm{n}}{2}\right) !}{\mathrm{n} !}\left(\mathrm{C} _{0}{ }^{2}-2 \mathrm{C} _{1}{ }^{2}+3 \mathrm{C} _{2}{ }^{2}-\ldots \ldots \ldots+(-1)^{\mathrm{n}}(\mathrm{n}+1) \mathrm{C} _{\mathrm{n}}{ }^{2}\right)$ when $\mathrm{n}$ is an even positive integer, is equal to

(a) $(-1)^{\mathrm{n} / 2}(\mathrm{n}+2)$

(b) $(-1)^{\mathrm{n}}(\mathrm{n}+1)$

(c) $(-1)^{\mathrm{n} / 2}(\mathrm{n}+1)$

(d) None of these

Show Answer Answer: a

2. $\left(\begin{array}{c}30 \ 0\end{array}\right)\left(\begin{array}{c}30 \ 10\end{array}\right)-\left(\begin{array}{c}30 \ 1\end{array}\right)\left(\begin{array}{c}30 \ 11\end{array}\right)+\ldots \ldots \ldots+\left(\begin{array}{l}30 \ 20\end{array}\right)\left(\begin{array}{l}30 \ 30\end{array}\right)$ is equal to

(a) ${ }^{30} \mathrm{C} _{11}$

(b) ${ }^{60} \mathrm{C} _{10}$

(c) ${ }^{30} \mathrm{C} _{10}$

(d) ${ }^{65} \mathrm{C} _{55}$

Show Answer Answer: c

3. If $\mathrm{r}=0,1,2$ 10 , let $\mathrm{A} _{r}, \mathrm{~B} _{r}$ and $\mathrm{C} _{r}$ denote, respectively, the coefficent of $\mathrm{x}^{r}$ in the expansions of $(1+x)^{10},(1+x)^{20}$ and $(1+x)^{30}$. Then $\sum\limits _{r=1}^{10} \mathrm{~A} _{\mathrm{r}}\left(\mathrm{B} _{10} \mathrm{~B} _{\mathrm{r}}-\mathrm{C} _{10} \mathrm{~A} _{\mathrm{r}}\right)$ is equal to

(a) $\mathrm{B} _{10}-\mathrm{C} _{10}$

(b) $\mathrm{A} _{10}\left(\mathrm{~B} _{10}^{2}-\mathrm{C} _{10} \mathrm{~A} _{10}\right)$

(c) 0

(d) $\mathrm{C} _{10}-\mathrm{B} _{10}$

Show Answer Answer: d

4. Value of $2^{k}\left(\begin{array}{l}n \ 0\end{array}\right)\left(\begin{array}{l}n \ k\end{array}\right)-2^{k-1}\left(\begin{array}{l}n \ 1\end{array}\right)\left(\begin{array}{l}n-1 \ k-1\end{array}\right)+2^{k-2}\left(\begin{array}{l}n \ 2\end{array}\right)\left(\begin{array}{l}n-2 \ k-2\end{array}\right)-\ldots \ldots+(-1)^{k}\left(\begin{array}{l}n \ k\end{array}\right)\left(\begin{array}{c}n-k \ 0\end{array}\right)$ is

(a) $\left(\begin{array}{l}\mathrm{n} \ \mathrm{k}\end{array}\right)$

(b) $\left(\begin{array}{l}n-1 \ \mathrm{k}-1\end{array}\right)$

(c) 1

(d) None of these

Show Answer Answer: a

5. $\sum\limits _{\mathrm{r}=0}^{\mathrm{n}}(-1)^{\mathrm{r}}{ }^{n} C _{\mathrm{r}}\left(\dfrac{1}{2^{\mathrm{r}}}+\dfrac{3^{\mathrm{r}}}{2^{2 \mathrm{r}}}+\dfrac{7^{\mathrm{r}}}{2^{3^{\mathrm{r}}}}+\dfrac{15^{\mathrm{r}}}{2^{4 \mathrm{r}}}+\ldots \ldots \ldots\right.$. up to $\mathrm{m}$ tems $\left.)\right)=$

(a) $\dfrac{2^{\mathrm{mn}}-1}{2^{\mathrm{mn}}\left(2^{\mathrm{n}}-1\right)}$

(b) $\dfrac{2^{m}-2^{n}}{m-n}$

(c) 1

(d) None of these

Show Answer Answer: a

6. If $\sum\limits _{\mathrm{r}=0}^{2 \mathrm{n}} \mathrm{a} _{\mathrm{r}}(\mathrm{x}-2)^{\mathrm{r}}=\sum\limits _{\mathrm{r}=0}^{2 \mathrm{n}} \mathrm{b} _{\mathrm{r}}(\mathrm{x}-3)^{\mathrm{r}}$ and $\mathrm{a} _{\mathrm{k}}=1$ for all $\mathrm{k} \geq \mathrm{n}$, then $\mathrm{b} _{\mathrm{n}}=$

(a) ${ }^{n} C _{n}$

(b) ${ }^{2 n+1} \mathrm{C} _{\mathrm{n}+1}$

(c) ${ }^{2 n+1} C _{n}$

(d) None of these

Show Answer Answer: b

7. If $(1+x)^{n}=C _{0}+C _{1} x+C _{2} x^{2}+\ldots \ldots \ldots+C _{n} x^{n}$, then the sum of the products of the $C _{i}{ }^{\prime} \mathrm{s}$ taken two at a time represented by $\sum \sum \mathrm{C} _{\mathrm{i}} \mathrm{C} _{\mathrm{j}}(0 \leq \mathrm{i}<\mathrm{j} \leq \mathrm{n})$ is equal to

(a) $2^{2 n-1}$

(b) $2^{\mathrm{n}}-\dfrac{(2 \mathrm{n}) !}{2(\mathrm{n} !)^{2}}$

(c) $2^{2 \mathrm{n}-1}-\dfrac{(2 \mathrm{n}) !}{2(\mathrm{n} !)^{2}}$

(d) None of these

Show Answer Answer: c

8. Given $\mathrm{s} _{\mathrm{n}}=1+\mathrm{q}^{2} \mathrm{q}^{2}+\ldots \ldots \ldots .+\mathrm{q}^{\mathrm{n}}$ and $\mathrm{S} _{\mathrm{n}}=1+\dfrac{\mathrm{q}+1}{2}+\left(\dfrac{\mathrm{q}+1}{2}\right)^{2}+\ldots \ldots \ldots+\left(\dfrac{\mathrm{q}+1}{2}\right)^{\mathrm{n}}, \mathrm{q} \neq 1$ then ${ }^{n+1} \mathrm{C} _{1}+{ }^{n+1} \mathrm{C} _{2} \mathrm{~S} _{1}+{ }^{n+1} \mathrm{C} _{3} \mathrm{~S} _{2}+\ldots \ldots . .+{ }^{\mathrm{n+1}} \mathrm{C} _{\mathrm{n}+1} \mathrm{~S} _{\mathrm{n}}=$

(a) $2^{\mathrm{n}} \mathrm{S} _{\mathrm{n}}$

(b) $\mathrm{S} _{\mathrm{n}}$

(c) $\dfrac{S _{n}}{2^{n}}$

(d) None of these

Show Answer Answer: a

9. $\lim _{n \rightarrow \infty} \sum\limits _{\mathrm{r}=0}^{\mathrm{n}}\left(\begin{array}{l}\mathrm{n} \ \mathrm{r}\end{array}\right) \dfrac{1}{(\mathrm{r}+3) \mathrm{n}^{\mathrm{r}}}=$

(a) e

(b) e-1

(c) $\mathrm{e}+1$

(d) e-2

Show Answer Answer: d

10. The coefficient of $x^{8}$ is the expansion of $\left(1+\dfrac{x^{2}}{2 !}+\dfrac{x^{4}}{4 !}+\dfrac{x^{6}}{6 !}+\dfrac{x^{8}}{8 !}\right)^{2}$ is

(a) $\dfrac{1}{315}$

(b) $\dfrac{2}{315}$

(c) $\dfrac{1}{105}$

(d) $\dfrac{1}{210}$

Show Answer Answer: a

11. Match the following:

Column I Column II
(a) The sum of binomial coefficients of terms containing power of $x$ more than $x^{20}$ in $(1+x)^{41}$ is divisibile by (p) $2^{39}$
(b) The sum of binomial coefficients of rational terms in the expansion of $(1+\sqrt{2})^{42}$ is divisible by (q) $2^{40}$
(c) If $\left(x+\dfrac{1}{x}+x^{2}+\dfrac{1}{x^{2}}\right)^{21}=a _{0} x^{-42}+a _{1} x^{-41}+a _{2} x^{-40}+\ldots \ldots . .+a _{82} x^{40}$, then $\mathrm{a} _{0}+\mathrm{a} _{2}+\ldots . .+\mathrm{a} _{82}$ is divisible by (r) $2^{41}$
(d) The sum of binomial coefficients of positive real terms in the expansion of $(1+\mathrm{ix})^{42}(\mathrm{x}>0)$ is divisible by (s) $2^{38}$
Show Answer Answer: $\mathrm{a} \rightarrow \mathrm{p}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{r}, \mathrm{d} \rightarrow \mathrm{q}$

12. Read the passage and answer the questions that follow:

An equation $\mathrm{a} _{0}+\mathrm{a} _{1} \mathrm{x}+\mathrm{a} _{2} \mathrm{x}^{2}+$ $+\mathrm{a} _{99} \mathrm{x}^{99}+\mathrm{x}^{100}=0$ has roots ${ }^{99} \mathrm{C} _{0},{ }^{99} \mathrm{C} _{1},{ } _{,}^{99} \mathrm{C} _{2}$ ${ }^{99} \mathrm{C} _{99}$.

(i) The value of $\mathrm{a} _{99}$ is

(a) $2^{98}$

(b) $2^{99}$

(c) $-2^{99}$

(d) None of these

(ii) The value of $\mathrm{a} _{98}$ is

(b) $\dfrac{2^{198}+{ }^{198} \mathrm{C} _{99}}{2}$

(c) $2{ }^{99}-{ }^{99} \mathrm{C} _{49}$

(d) None of these

(a) $\dfrac{2^{198}-{ }^{198} \mathrm{C} _{99}}{2}$

(iii) The value of $\left({ }^{99} \mathrm{C} _0\right)^2+\left({ }^{99} \mathrm{C} _1\right)^2+\ldots \ldots \ldots \ldots+\left({ }^{99} \mathrm{C} _{99}\right)^2 \text { is }$

(a) $2 \mathrm{a} _{98}-\mathrm{a} _{99}^{2}$

(b) $\mathrm{a} _{98}^{2}-\mathrm{a} _{99}^{2}$

(c) $\mathrm{a} _{99}^{2}-2 \mathrm{a} _{98}$

(d) None of these

Show Answer Answer: (i) c (ii) a (iii) c


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