For $n \in \mathbb{N}$,

$(a+b)^n ={ }^n C_0 a^n+{ }^n C_1 a^{n-1} b+\cdots+{ }^n C_{n-1} a b^{n-1} +{ }^n C_n b^n$

$=\sum_{k=0}^n{ }^n C_k a^{n-k} b^k$

$(a-b)^n =\sum_{k=0}^n{ }^n C_k(-1)^k a^{n-k} b^k$

1) For $n \in \mathbb{N}$, 0 $\leqslant$ r < n, we have

${ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r$

2) $(a+b)^n+(a-b)^n=2 \sum_{\substack{k=0 \ k \text { even }}}^n{ }^n C_k a^{n-k} b^k$

3) $(a+b)^n-(a-b)^n =2 \sum_{\substack{k=0 \ k \text { odd }}}^n{ }^n C_k a^{n-k} b^k$

For $m, n \in \mathbb{N}, k \geqslant 0$

$\sum_{r=0}^k{ }^n C_r{ }^m C_{k-r}$ is the coefficient $\text { of } x^k \text { in }(1+x)^n(1+x)^m.$

Which one among $101^{50}$ and $99^{50}+100^{50}$ is bigger?

Solution:

$101^{50} = (100+1)^{50}$

$99^{50}=(100-1)^{50}$

$101^{50}-99^{50}=(100+1)^{50}-(100-1)^{50}$

$=\sum_{k=0}^{50}{ }^{50} C_k(100)^{50-k}-\sum_{k=0}^{50}{ }^{50} C_k(100)(-1)$

$=2 \sum_{k=0}^{50}{ }^{50} C_k(100)^{50-k} \ \ (k \ \text{odd})$

$2({ }^{50} C_1 100^{49}+{ }^{50} C_3 100^{47}+ …. +{ }^{50} C_{49} 100)$

$\underbrace{50} \ \ \{\because { }^{50} C_1 =50 \}$

$=100^{50}+2^{50} C_3 100^{47}+\cdots+2^{50} C_{49} 100$

$101^{50}-99^{50}>100^{50}$

$\Rightarrow 101^{50}>100^{50}+99^{50}$

If for a positive integer $n$, the coefficients of $2 n d, 3 r d$ and $4 t h$ term in the expansion of $(1+x)^n$ are in arithmetic progression, then what is $n$ ?

Solution:

$(1+x)^n$=

${ }^n C_0+{ }^n C_1 x+{ }^n C_2 x^2+{ }^n C_3 x^3+\cdots+{ }^n C_n x^n$

$\text { 2nd term } \longrightarrow{ }^n C_1=(n)$

$\text { 3rd term } \longrightarrow{ }^n C_2=\left(\frac{n(n-1)}{2}\right)$

$\text {4th term } \rightarrow{ }^n C_3=\left(\frac{n(n-1)(n-2)}{6}\right.$

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

$6+(n-1)(n-2)=6 n-6$

$\Rightarrow 6+n^2-n-2 n+2=6 n-6$

$\Rightarrow n^2-9 n+14=0$

$n=\frac{9 \pm \sqrt{81-56}}{2} =\frac{9 \pm 5}{2}$

$n= \underbrace{2}_{\text{Not possible}} \text { or } 7$

$\biggl\{ (1+x)^n \ \ \text { there are total $ n+1$ terms } \\ \text{If} \ n=2 , \ \text{we get 3 terms}\biggl \}$

What is the sum of all the rational numbers in the expansion of $\left(\sqrt{2}+3^{1 / 5}\right)^{10}?$

Solution: $\left(\sqrt{2}+3^{1 / 5}\right)^{10}$

${ }^{10} C_{0}(\sqrt{2})^{10}(3^{\frac{1}{5}})^{0} + { }^{10} C_{1}(\sqrt{2})^9(3^{\frac{1}{5}})^{1}$

$+ …. + { }^{10} C_{10}(\sqrt{2})^0(3^{\frac{1}{5}})^{10}$ (1.1.9=9)

$1.32.1=32$

$39+9=41$

The expression $\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5$ is a polynomial of degree

$(1) 5 \quad$ $(2) 6 \quad$ $(3) 7 \quad$ $(4) 8$

Solution:

$\left(x+\sqrt{x^3-1}\right)^5=\sum_{k=0}^5 5 C_k x^{5-k}\left(\sqrt{x^3-1}\right)^k$

$\left(x-\sqrt{x^3-1}\right)^5=\sum_{k=0}^5 5 C_k x^{5-k}(-1)^k\left(\sqrt{x^3-1}\right)^k$

$\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5=2 \sum_{k=0}^5 C_k x^{5-k}\left(x^3-1\right)^{k / 2}$ k even

$[k=0 \quad x^5 \quad \text {1 degree 5} ]$

$[k=2 \quad x^3\left(x^3-1\right) \text { degree 6} ]$

$[k=4 \quad x(x^3-1)^2 \text{ degree 7} ]$

The term independent of $x$ in the expansion of $(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1} - \frac{x-1}{x-x^{1 / 2}})^{10}$ is

$(1) 4$

$(2) 120$

$(3) 210$

$(4) 310$

$\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}=\frac{\left(x^{1 / 3}\right)^3+1}{x^{2 / 3}-x^{1 / 3}+1}=\frac{\left(x^{1 / 3}+1\right)\left(x^{2 / 3}-x^{1 / 3}+1\right)}{\left(x^{2 / 3}-x^{1 / 3}+1\right)}$

$\frac{x-1}{x-x^{1 / 2}}=\frac{\left(x^{1 / 2}\right)^2-1}{x^{1 / 2}\left(x^{1 / 2}-1\right)}=\frac{\left(x^{1 / 2}+1\right)\left(x^{1 / 2}-1\right)}{x^{1 / 2}\left(x^{1 / 2}-1\right)}=\frac{x^{1 / 2}+1}{x^{1 / 2}}$

$=1+x^{-1 / 2}$

$\therefore (\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1} - \frac{x-1}{x-x^{1 / 2}})^{10} =(x^{1 / 3}+1-1-x^{-1 / 2})^{10}$

$=\left(x^{1 / 3}-x^{-1 / 2}\right)^{10}$

$=\sum_{k=0}^{10}{ }^{10} C_k\left(x^{1 / 3}\right)^{10-k}(-1)^k x^{-k / 2}$

$=\sum_{k=0}^{10}(-1)^k \quad { }^{10} C_k x^{(20-5 k) / 6}$

$\frac{10-k}{3}-\frac{k}{2}$

$=\frac{20-2 k-3 k}{6}$

$=\frac{20-5 k}{6}$

$\frac{20-5 k}{6}=0$

$\Rightarrow k=4$

$(-1)^4$ ${ }^{10} C_4$

$=1 \cdot \frac{10 \cdot 9 \cdot 8 \cdot 7}{24}=210$

The value of the sum ${ }^{50} C_4+\sum_{r=1}^6{ }^{56-r} C_3$ is

$(1) { }^{55} C_4$

$(1) { }^{55} C_3$

$(1) { }^{56} C_3$

$(1) { }^{56} C_4$

$n \in \mathbb{N}$, $\ \ 0 \leqslant r < n$,

We have,

${ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r$

${ }^{50} C_4+{ }^{50} C_3+{ }^{51} C_3+{ }^{52} C_3+{ }^{53} C_3+{ }^{54} C_3+{ }^{55} C_3$

$={ }^{51} C_4+{ }^{51} C_3+{ }^{52} C_3+{ }^{53} C_3+{ }^{54} C_3+{ }^{55} C_3$

$={ }^{52} C_4+{ }^{52} C_3+{ }^{53} C_3+{ }^{54} C_3+{ }^{55} C_3$

$={ }^{53} C_4+{ }^{53} C_3+{ }^{54} C_3+{ }^{55} C_3$

$={ }^{54} C_4+{ }^{54} C_3+{ }^{55} C_3$

$={ }^{55} C_4+{ }^{55} C_3$

$={ }^{56} C_4$

The value of $\left({ }^{21} C_1-{ }^{10} C_1\right)+\left({ }^{21} C_2-{ }^{10} C_2\right)+\cdots+\left({ }^{21} C_{10}-{ }^{10} C_{10}\right)$ is

(1) $2^{20}-2^{10}$

(2) $2^{21}-2^{11}$

(3) $2^{21}-2^{10}$

(4) $2^{20}-2^9$

Solution:

$\left({ }^{21} C_1+{ }^{21} C_2+\cdots+{ }^{21} C_{10}\right)-\underbrace {\left({ }^{10} C_1+{ }^{10} C_2+\cdots+{ }^{10} C_{10}\right)}_{2^{10}-1}$

$\sum_{k=0}^{10} { }^{10} C_k(1)^{10-k}(1)^k-{ }^{10} C_0 =(1+1)^{10}-1=2^{10}-1$

${ }^{21} C_1+{ }^{21} C_2+\cdots+{ }^{21} C_{10}$

$=\frac{1}{2}\left(2^{21} C_1+2^{21} C_2+\cdots+2^{21} C_{10}\right)$

$=\frac{1}{2}({{ }^{21} C_1+{ }^{21} C_2+\cdots+{ }^{21} C_{20}})$

${{ }^{21} C_0+{ }^{21} C_{21}-{ }^{21} C_0+ { }^{21} C_{21}})$

${ }^{21} C_1={ }^{21} C_{20}$

${ }^{21} C_2={ }^{21} C_{19}$

$.$

$.$

$.$

${ }^{21} C_{10}={ }^{21} C_{11}$

$(1+1)^{21} =\frac{1}{2}\left(2^{21}-1-1\right) =2^{20}-1$

The value of ${ }^{20} C_0-{ }^{20} C_1+{ }^{20} C_2-\cdots+{ }^{20} C_{10}$ is

(1) 0

(2) ${ }^{20} C_{10}$

(3) $-{ }^{20} C_{10}$

(4) $\frac{{ }^{20} C_{10}}{2}$

$Y={ }^{20} C_0-{ }^{20} C_1+{ }^{20} C_2-\cdots+{ }^{20} C_{10}$

$(1-x)^{20}={ }^{20} C_0-{ }^{20} C_1 x+{ }^{20} C_2 x^2-\cdots$

$+{ }^{20} C_{10} x^{10}-20 C_{11} x^{11}+\cdots+{ }^{20} C_{20} x^{20}$

$\text { Put } x =1$

$0 =Y-{ }^{20} C_{11}+{ }^{20} C_{12}-\cdots+{ }^{20} C_{20}$

$Y={ }^{20} C_9-{ }^{20} C_8+{ }^{20} C_7-\cdots-{ }^{20} C_0 +{ }^{20} C_{10}-{ }^{20} C_{10}$

= $-{\left({ }^{20} C_{10}-{ }^{20} C_9+{ }^{20} C_8-\cdots+{ }^{20} C_0\right)}{+{ }^{20} C_{10}}$

$2 Y={ }^{20} C_{10}$

$Y=\frac{1}{2}{ }^{20} C_{10}$

The value of the sum ${ }^{30} C_0{ }^{30} C_{10}-{ }^{30} C_1{ }^{30} C_{11}+{ }^{30} C_2{ }^{30} C_{12}-\cdots+{ }^{30} C_{20}{ }^{30} C_{30}$ is

(1) ${ }^{30} C_{10}$

(2) ${ }^{30} C_{15}$

(3) ${ }^{60} C_{30}$

(4) ${ }^{31} C_{10}$

${ }^{30} C_0{ }^{30} C_{20}-{ }^{30} C_1{ }^{30} C_{19}+{ }^{30} C_2{ }^{30} C_{18}-\cdots+{ }^{30} C_{20}{ }^{30} C_0$

co-efficient of $x^{20}$ in $(1+x)^{30}(1-x)^{30}$

$(1+x)^{30}(1-x)^{30}$

$=\left(1-x^2\right)^{30}$

$=\sum_{k=0}^{30}{ }^{30} C_k(-1)^k x^{2 k}$

$\text { co-efficient of } x^{20}={ }^{30} C_{10}(-1)^{10}$ $={ }^{30} C_{10}$

The sum of the coefficients of the integral powers of $x$ in the binomial expansion of $(1-2 \sqrt{x})^{50}$ is

(1) $\frac{3^{50}-1}{2}$

(2) $\frac{3^{50}+1}{2}$

(3) $\frac{2^{50}+1}{2}$

(4) $\frac{3^{50}}{2}$

$(1-2 \sqrt{x})^{50} = \sum_{k=0}^{50}{ }^{50} C_k(-1)^k 2^k x{ }^{k / 2}$

${ }^{50} C_0+{ }^{50} C_2 2^2+{ }^{50} C_4 2^4+\cdots+{ }^{50} C_{50} 2^{50}$

$(1+2 x)^{50}+(1-2 x)^{50}$

$= \sum_{k=0}^{50}{ }^{50} C_k 2^k x^k+\sum_{k=0}^{50}{ }^{50} C_k(-1)^k 2^k x^k$

$= 2 \sum_{k=0}^{50}{ }^{50} C_k 2^k x^k \ \ k= \text{even}$

Putting $x=1$,

${ }^{50} C_0+{ }^{50} C_2 2^2+{ }^{50} C_4 2^4+\cdots+{ }^{50} C_{50} 2^{50}$

$=\frac{1}{2}\left\{3^{50}+(-1)^{50}\right\}$

$=\frac{3^{50}+1}{2}$