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Q. $\displaystyle \sum _{r = 0}^{n} \left(\frac{r^{2}}{r + 1}\right)^{n}C_{r}$ is equal to

NTA AbhyasNTA Abhyas 2020Binomial Theorem

Solution:

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