Question

Jee Question
If
$\sf \: \: (1 + x) ^{2} = C_{0} + C_1x + C_2 {x}^{2} + C_3 {x}^{3} + .... + C_nx {}^{n}$
Show that
$\displaystyle \sum^{ \sf \: n} _{ \sf \: r = 0} \sf \: \dfrac{C_r {3}^{r + 4} }{(r + 1)(r + 2)(r + 3)(r + 4)} = \dfrac{1}{(n + 1)(n + 2)(n + 3)(n + 4)} \bigg( {4}^{n + 4} - \displaystyle \sum^{ \sf \: 3} _{ \sf \: t= 0} \sf \: \:^{n + 4} C_t \: 3^{t} \bigg)$
​

Answer 1

Consider the RHS.

$\displaystyle\longrightarrow S=\dfrac{1}{(n+1)(n+2)(n+3)(n+4)}\left(4^{n+4}-\sum_{t=0}^3\,^{n+4}C_t\,3^t\right)$

We know that,

$\longrightarrow (1+x)^n=\,^n\!C_0+\,^n\!C_1\,x+\,^n\!C_2\,x^2+\,^n\!C_3\,x^3+\,\dots\,+\,^n\!C_n\,x^n$

or,

$\displaystyle\longrightarrow (1+x)^n=\sum_{t=0}^n\,^n\!C_t\,x^t$

Taking $x=3$ and replacing $n$ by $n+4$ we get,

$\displaystyle\longrightarrow4^{n+4}=\sum_{t=0}^{n+4}\,^{n+4}C_t\,3^t$

Then our sum becomes,

$\displaystyle\longrightarrow S=\dfrac{1}{(n+1)(n+2)(n+3)(n+4)}\left(\sum_{t=0}^{n+4}\,^{n+4}C_t\,3^t-\sum_{t=0}^3\,^{n+4}C_t\,3^t\right)$

$\displaystyle\longrightarrow S=\dfrac{1}{(n+1)(n+2)(n+3)(n+4)}\sum_{t=4}^{n+4}\,^{n+4}C_t\,3^t$

Put,

• $r=t-4$

• $t=r+4$

• $t=4\quad\implies\quad r=0$

• $t=n+4\quad\implies\quad r=n$

Then our sum becomes,

$\displaystyle\longrightarrow S=\dfrac{1}{(n+1)(n+2)(n+3)(n+4)}\sum_{r=0}^n\,^{n+4}C_{r+4}\,3^{r+4}$

Taking $^{n+4}C_{r+4}=\dfrac{(n+4)!}{(r+4)(n-r)!},$

$\displaystyle\longrightarrow S=\dfrac{1}{(n+1)(n+2)(n+3)(n+4)}\sum_{r=0}^n\dfrac{(n+4)!}{(r+4)!(n-r)!}\,3^{r+4}$

Take,

• $(n+4)!=n!(n+1)(n+2)(n+3)(n+4)$

• $(r+4)!=r!(r+1)(r+2)(r+3)(r+4)$

Then,

$\displaystyle\footnotesize\longrightarrow S=\dfrac{1}{(n+1)(n+2)(n+3)(n+4)}\sum_{r=0}^n\dfrac{n!(n+1)(n+2)(n+3)(n+4)}{r!(r+1)(r+2)(r+3)(r+4)(n-r)!}\,3^{r+4}$

$\displaystyle\footnotesize\longrightarrow S=\dfrac{(n+1)(n+2)(n+3)(n+4)}{(n+1)(n+2)(n+3)(n+4)}\sum_{r=0}^n\dfrac{n!}{r!(r+1)(r+2)(r+3)(r+4)(n-r)!}\,3^{r+4}$

$\displaystyle\longrightarrow S=\sum_{r=0}^n\dfrac{n!}{r!(r+1)(r+2)(r+3)(r+4)(n-r)!}\,3^{r+4}$

$\displaystyle\longrightarrow S=\sum_{r=0}^n\dfrac{1}{(r+1)(r+2)(r+3)(r+4)}\cdot\dfrac{n!}{r!(n-r)!}\,3^{r+4}$

$\displaystyle\longrightarrow S=\sum_{r=0}^n\dfrac{1}{(r+1)(r+2)(r+3)(r+4)}\cdot\,^nC_r\,3^{r+4}$

Taking $^nC_r=C_r$ we get the LHS.

$\displaystyle\longrightarrow S=\sum_{r=0}^n\dfrac{C_r\,3^{r+4}}{(r+1)(r+2)(r+3)(r+4)}$

Hence Proved!