$(x+y)^n={ }^n c_0 x^n+{ }^n C_1 x^{n-1} y + { }^n C_2 x^{n-2} y^2+{ }^n C_3 x^{n-3} y^3 + \cdots +{ }^n C_{n-1} x y^{n-1}+{ }^n C_n y^n$

$\biggl [{ }^n C_r=\frac{n !}{r !(n-r) !}\biggl]$

In $(a - b)^n \ (n \geqslant 5),$ the sum of $5^{\text{th}}$ and $6^\text{{th}}$ terms is $0.$ Then find the value of $\frac{a}{b}.$

Solution:

$(a-b)^n={ }^n C_0 a^n- { }^n C_1 a^{n-1} b+\quad+ \underbrace{{ }^n C_4 a^{n-4} b^4} _{5^{\text{th}} \ \text{term}} - \underbrace{{ }^n C _5 a^{n-5} b^5} _{6^{\text{th}} \ \text{term}}$
Now, according to the question,

${ }^n C_4 a^{n-4} b^4={ }^n C_5 a^{n-5} b^5$

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

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

$\left(1+x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\cdots a_{2 n} x^{2 n}$

1) $a_0+a_1+a_2+\ldots+a_{2 n}=$ ?

2) $a_0-a_1+a_2-a_3+a_4-\ldots+a_{2 n}=$ ?

3) $a_0^2-a_1^2+a_2^2-a_3^2+\cdots+a_{2 n}^2=$ ?

Solution:

1) $a_0+a_1+a_2+\ldots+a_{2 n}= 3^n$
2) $a_0-a_1+a_2-a_3+a_4-\ldots+a_{2 n}= 1$
3) $(1+x+x^2)^n (1-\frac{1}{x}+\frac{1}{x^2})^n$

$a_0+a_1 x+a_2 x^2+\cdots +a_{2 n} x^{2 n}$

$a_0-\frac{a_1}{x}+\frac{a_2}{x^2}-\quad+a_{2 n} / x^{2 n}$

term independent of $x$ is $a_0^2-a_1^2+a_2^2-a_3^2+\cdots \cdot .$

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

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

$= [1- \frac{1}{x} + \frac{1}{x^2} + x -1 + \frac{1}{x}+ x^2 - x +1]^n$

$=\left[\frac{1}{x^2}+1+x^2\right]^n$

$=\frac{1}{x^{2 n}}\left(1+x^2+x^4\right)^n$

$=\frac{1}{x^{2 n}}\left(a_0+a_1 x^2+a_2 x^4+\cdots a_{n} x^{2 n} \cdots a_{2 n} x^{4 n}\right)$

$\Rightarrow$ term independent of $x$ is $a_n$.

$(1+x)^{14}$ if the $r^{\text {th }}, r+1^{\text {th }}, r+2^{\text {th }}$ terms are in Arithmetic Progression (AP). $r=$ ?

Solution:

$r\text { th } \rightarrow 14-r+1$ $\rightarrow \quad { }^{14} c_{r-1} \cdot x^{r-1}$

$r+1 \text { th } \rightarrow \quad{ }^{14} \mathrm{C}_r x^r$

$r+2 \text { th } \rightarrow \quad{ }^{14} c_{r+1} x^{r+1}$

$\frac{14 !}{(r-1) !(15-r) !} \cdot x^{r-1} +\frac{14 !}{(r+1) !(13-r) !} \cdot x^{r+1}=2 \frac{14 !}{(4-r) ! r !} x^r$
$\frac{1}{(14-r)(15-r)} +\frac{x^2}{r(r+1)}=\frac{2 x}{r(14-r)}$

$(1+x)^n=\frac{n !}{0 ! n !}+\frac{n !}{1 !(n-1) !} x +\frac{n !}{2 !(n-2) !} x^2+\cdots$

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

$\frac{n(n-1)(n-2)(n-3)}{4 !}+\cdots \cdot \ \ \$ $|x| < 1$

$\lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)=1$

$\lim _{x \rightarrow 0} e^x-1=\lim _{x \rightarrow 0} 1+x$

$e =\lim _{x \rightarrow 0}(1+x)^{1 / x}$

$=2 \cdot 7182818 \ldots$

$=\lim _{y \rightarrow \infty}(1+1 / y)^y$

$e^x=\lim _{y \rightarrow \infty}(1+1 / y)^{x y}$

$=\lim _{y \rightarrow \infty}\left[1+x y+\frac{x y(x y-1)}{2 !} \cdot \frac{1}{y^2} + \frac{x y(x y - 1)(x y-2)}{3 ! y^3} +\ldots \right]$

$e^x=\lim _{y \rightarrow \infty}\left[1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\frac{x^4}{4 !}+\cdots\right]$

$e^{-x}=\left[1-x+\frac{x^2}{2 !}-\frac{x^3}{3 !}+\frac{x^4}{4 !}+\cdots \cdot\right]$

$e=1+1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\cdots$

Prove that $C_1+2 C_2+3 C_3+\cdots+n C_n=n \cdot 2^{n-1}$

Solution:
$C_0+c_1+C_2+C_3+\cdots \quad C_n=2^n$

$\frac{d}{d x} x^n=n x^{n-1}$

$(1+x)^n=C_0+C_1 x+C_2 x^2+\cdots \cdot C_n x$

$n(1+x)^{n-1}=C_1+2 C_2 x+3 C_3 x^2+\cdots+n C_n x^{n-1}$

Plug in$x=1$

$\Rightarrow n \cdot 2^{n-1}=C_1+2 C_2+3 C_3+\ldots+n C_n$

$C_0+2 C_1+3 C_2+\cdots+(n+1) C_n=(n+2) 2^{n-1}$

Solution:

L.H.S $= \left(C_0+C_1+C_2+\cdots C_n\right) +\left(C_1+2 C_2+3 C_3+\cdots+n C_n\right) \\ = 2^n +n \cdot 2^{n-1}$

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

Other Way to prove,

$C_0+2 C_1+3 c_2+\cdots+(n+1) C_n=k$

$(n+1) C_n+n C_{n-1}+(n-1) C_{n-2}+\ldots+C_0=k$

$(n+2) C_0+(n+2) C_1+(n+2) C_2+\cdots(n+2) C_n=2 k$

$k=\frac{n+2}{2}\left(C_0+C_1+C_2+\ldots+C_n\right)=\frac{n+2}{2} \cdot 2^n$