Question

If $\left(1 + x\right)^{n}=C_{0}+C_{1}x+C_{2}x^{2}+....+C_{n}x^{n}$ and $\displaystyle \sum _{r = 0}^{50} \frac{C_{r}^{2}}{\left(r + 1\right)}=\frac{m !}{\left(n !\right)^{2}}$ , 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)

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