Question

If ${\left(1+x\right)}^n=C_0+C_{1}x+C_2{x}^2+....+C_n{x}^n$ and $\sum _{r=0}^{50}\frac{C_r^2}{\left(r+1\right)}=\frac{m!}{{\left(n!\right)}^2}$ , then the value of $\left(m+n\right)$ is equal to (where $C_r$ represents $\_{}^n{C}_r^{}$ )

• A. $149$
• B. $152$
• C. $155$
• D. $146$

Answer 1

Answer: (B)

${\left(1+x\right)}^{50}=C_0+C_{1}x+C_2{x}^2+.....+C_{50}{x}^{50}=\sum _{r=0}^{50}{C}_r{x}^r$
${\int }_0^x{\left(1+x\right)}^{50}dx=\sum _{r=0}^{50}{\int }_0^x{C}_r{x}^{r}dx$
$\frac{{\left(1+x\right)}^{51}-1}{51}=\sum _{r=0}^{50}\frac{C_r{x}^{r+1}}{r+1}$
and ${\left(x+1\right)}^{50}=\sum _{r=0}^{50}{C}_r{x}^{50-r}$
Multiplying these two equations, we get,
$\left(\frac{{\left(1+x\right)}^{51}-1}{51}\right){\left(x+1\right)}^{50}=\left(\sum _{r=0}^{50}\frac{C_r{x}^{r+1}}{r+1}\right)\left(\sum _{r=0}^{50}{C}_r{x}^{50-r}\right)$
$\frac{{\left(1+x\right)}^{101}-{\left(1+x\right)}^{50}}{51}=$ $\left(\frac{C_{0}x}{1}+\frac{C_1{x}^2}{2}+\frac{C_2{x}^3}{3}+....+\frac{C_{50}{x}^{51}}{51}\right)\left(C_0{x}^{50}+C_1{x}^{49}+C_2{x}^{48}+....+C_{50}\right)$
Now, comparing the coefficient of $x^{51},$ we get,
$\frac{1}{51}\left({}^{101}{C}_{51}-0\right)=\frac{C_0^2}{1}+\frac{C_1^2}{2}+\frac{C_2^2}{3}+...+\frac{C_{50}^2}{51}=\sum _{r=0}^{50}\frac{C_r^2}{r+1}$
$\Rightarrow \sum _{r=0}^{50}\frac{C_r^2}{r+1}=\frac{1}{51}\times \frac{101!}{51!50!}=\frac{101!}{{\left(51!\right)}^2}=\frac{m!}{{\left(n!\right)}^2}$
$\Rightarrow m+n=101+51=152$