Question

If $c_0,c_1,c_2,......c_n$ are the coefficients in the expansion of $(1+x{)}^n$ , when $n$ is a positive integer, prove that
(1) $c_0-c_1+c_2-c_3+.......+(-1{)}^r{c}_r=(-1{)}^r\frac{|\underset{–}{n-1}}{|\underset{–}{r}|\underset{–}{n-r-1}}$
(2) $c_0-2c_1+3c_2-4c_3+.......+(-1{)}^n(n+1)c_n=0$
(3) $c_0^2-c_1^2+c_2^2-c_3^2+.......+(-1{)}^n{c}_n^2=0,$ or $(-1{)}^{\frac{n}{2}}{c}_{\frac{n}{2}},$
according as $n$ is odd or even.

Answer 1

${(1+x)}^n=1+{}^n{C}_{1}x+{}^n{C}_2{x}^2+...{.}^n{C}_n{x}^n$

$\Rightarrow {(x-1)}^n=1-{}^n{C}_{1}x+{}^n{C}_2{x}^2+....(-1{)}^n{C}_n{x}^n$

$\Rightarrow {(1+x)}^{n-1}=1{+}^{n-1}{C}_{1}x{+}^{n-1}{C}_2{x}^2+...{.}^{n-1}{C}_n{x}^{n-1}$

${1.}^n{C}_0{-}^n{C}_{1}x{+}^n{C}_2{x}^2+....{(-1)}^r{}^n{C}_r{x}^r$

$=$ Co efficient of $x^r$ in${(1-x)}^{n-1}$

$={}^{n-1}{C}_r.{(-1)}^r$

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

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

$=(-1)n\Sigma {(-1)}^{r-1}{.}^{n-1}{C}_{r-1}+0$ $=0+0=0$

$3.{(1+x)}^n{=}^n{C}_0{+}^n{C}_{1}x+...{+}^n{C}_n{x}^n⟶1$

$\Rightarrow {(x-1)}^n{=}^n{C}_0{x}^n{-}^n{C}_1{x}^n+...+{(-1)}^n{}^n{C}_n⟶2$

Multiplying 1 and 2:

$S=C_0^2-C_1^2+C_2^2+....+{(-1)}^n{C}_n^2$

$=$Co efficient of, $x^n$ in ${(x^2-1)}^n$

$\Rightarrow T_{r+1}{=}^n{C}_r{x}^{2(n-r)}{(-1)}^r$

$\Rightarrow 2n-2r=n$

$\Rightarrow r=\frac{n}{2}$

Therefore, For n = odd $\to$ r is invalid $\Rightarrow S=0$

For n = even $\to$ $S={(-1)}^{n/2}{}^n{C}_{n/2}$.