Question

If ${\sum }_{r=0}^n\frac{(-1{)}^r\cdot C_r}{(r+1)(r+2)(r+3)}=\frac{1}{a(n+b)},$ then $a+b$ is :

Answer 1

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

We know that $C_r=\frac{n!}{r!(n-r)!}$

Moreover, we know that $(r+1)\times r!=(r+1)!$

Multiply by $(n+1)(n+2)(n+3)$ in both numerator and denominator in $X$. We get :

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

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

$\frac{1}{(n+1)(n+2)(n+3)}\text{will come out of summation because it is independent of r.}$

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

We know that ${\sum }_{r=0}^n(-1{)}^r\cdot C_r^n=0$

Thus, ${\sum }_{r=-3}^n(-1{)}^r\cdot C_{r+3}^{n+3}=0$

${\sum }_{r=0}^n(-1{)}^r\cdot C_{r+3}^{n+3}+C_0^{n+3}-C_1^{n+3}+C_2^{n+3}=0$

${\sum }_{r=0}^n(-1{)}^r\cdot C_{r+3}^{n+3}=-(C_0^{n+3}-C_1^{n+3}+C_2^{n+3})$

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

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

Using the above result in $X$, we get:

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

Therefore, $X=\frac{1}{(-2)(n+3)}$

Comparing X with $\frac{1}{a(n+b)}$ we get :

$a=-2$ and $b=3$

Therefore, $a+b=-1+3=1$