Question

$\frac{3!}{2(n+3)}={\sum }_{r=0}^n(-1{)}^r\left(\frac{{}^n{C}_r}{{}^{r+3}{C}_r}\right)$

• A. True
• B. False

Answer 1

The correct option is A True

$RHS={\sum }_{r=0}^n{(-1)}^r\left(\frac{{n_C}_r}{{{r+3}_C}_r}\right)$

$={\sum }_{r=0}^n{(-1)}^r\left(\frac{\left(\frac{n!}{r!(n-r)!}\right)}{\left(\frac{(r+3)!}{r!3!}\right)}\right)$

$={\sum }_{r=0}^n{(-1)}^r(\frac{n!}{r!(n-r)!}\times \frac{r!3!}{(r+3)!})$

$=n!.3!{\sum }_{r=0}^n\left(\frac{{(-1)}^r}{(n-r)!(r+3)!}\right)$

$=\frac{n!.3!}{(n+3)(n+2)(n+1)}{\sum }_{r=0}^n\left(\frac{{(-1)}^r(n+3)(n+2)(n+1)}{(n-r)!(r+3)!}\right)$

$=\frac{3!}{(n+3)(n+2)(n+1)}{\sum }_{r=0}^n\left(\frac{{(-1)}^r(n+3)(n+2)(n+1)n!}{(n-r)!(r+3)!}\right)$

$=\frac{3!}{(n+3)(n+2)(n+1)}{\sum }_{r=0}^n{(-1)}^r\frac{(n+3)!}{(n-r)!(r+3)!}$

$=\frac{3!}{(n+3)(n+2)(n+1)}{\sum }_{r=0}^n{(-1)}^r{{(n+3)}_C}_{(r+3)}$

$=\frac{3!}{{(-1)}^3(n+3)(n+2)(n+1)}{\sum }_{r=0}^n{(-1)}^r{(-1)}^3{{(n+3)}_C}_{(r+3)}$

$=\frac{-3!}{(n+3)(n+2)(n+1)}{\sum }_{r=0}^n{(-1)}^{r+3}{{\left(n+3\right)}_C}_{(r+3)}$

$=\frac{-3!}{(n+3)(n+2)(n+1)}[{(-1)}^3{{(n+3)}_C}_3+{(-1)}^4{{\left(n+3\right)}_C}_4+{(-1)}^5{{\left(n+3\right)}_C}_5+----------+{(-1)}^{n+3}{{\left(n+3\right)}_C}_{(n+3)}]$

$=\frac{-3!}{(n+3)(n+2)(n+1)}[-{{(n+3)}_C}_3+{{\left(n+3\right)}_C}_4-{{\left(n+3\right)}_C}_5+----------+{(-1)}^{n+3}{{\left(n+3\right)}_C}_{(n+3)}]$

$=\frac{-3!}{(n+3)(n+2)(n+1)}[{{(n+3)}_C}_0-{{\left(n+3\right)}_C}_1+{{\left(n+3\right)}_C}_2-{{\left(n+3\right)}_C}_3+{{\left(n+3\right)}_C}_4-{{\left(n+3\right)}_C}_5+----------+{(-1)}^{n+3}{{\left(n+3\right)}_C}_{(n+3)}]-{{(n+3)}_C}_0+{{\left(n+3\right)}_C}_1-{{\left(n+3\right)}_C}_2$

$=\frac{-3!}{(n+3)(n+2)(n+1)}\left[{(1-1)}^{n+3}\right]-\frac{(n+3)!}{0!(n+3)!}+\frac{\left(n+3\right)!}{1!\left(n+2\right)!}-\frac{\left(n+3\right)!}{2!\left(n+1\right)!}$

$=\frac{-3!}{(n+3)(n+2)(n+1)}[0-1+\frac{\left(n+3\right)\left(n+2\right)!}{\left(n+2\right)!}-\frac{\left(n+3\right)\left(n+2\right)\left(n+1\right)!}{2\left(n+1\right)!}]$

$=\frac{-3!}{(n+3)(n+2)(n+1)}[0-1+\left(n+3\right)-\frac{\left(n+3\right)\left(n+2\right)}{2}]$

$=\frac{3!}{(n+3)(n+2)(n+1)}[1-\left(n+3\right)+\frac{\left(n+3\right)\left(n+2\right)}{2}]$

$=\frac{3!}{(n+3)(n+2)(n+1)}[1-n-3+\frac{\left(n+3\right)\left(n+2\right)}{2}]$

$=\frac{3!}{(n+3)(n+2)(n+1)}\left[\frac{2(1-n-3)+\left(n+3\right)\left(n+2\right)}{2}\right]$

$=\frac{3!}{(n+3)(n+2)(n+1)}\left[\frac{2-2n-6+n^2+5n+6}{2}\right]$

$=\frac{3!}{(n+3)(n+2)(n+1)}\left[\frac{n^2+3n+2}{2}\right]$

$=\frac{3!}{(n+3)(n+2)(n+1)}\left[\frac{(n+1)(n+2)}{2}\right]$

$=\frac{3!}{2(n+3)}$

$=LHS$