Jee QuestionIfShow that

1.	Jee QuestionIfShow that
Answer» 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, $\displaystyle\footnotesize\text{$\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}$}$ $\displaystyle\footnotesize\text{$\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!

Answer»

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,

$\displaystyle\footnotesize\text{$\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}$}$

$\displaystyle\footnotesize\text{$\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}$}$