If $\sum_{r=1}^{30}\frac{r^2\left({}^{30}C_r ight)^2}{{}^{30}C_{r-1}}=\alpha imes2^{29}$, then $\alpha$ is equal to _______

Question

If $\sum_{r=1}^{30}\frac{r^2\left({}^{30}C_right)^2}{{}^{30}C_{r-1}}=\alpha	imes2^{29}$, then $\alpha$ is equal to _______

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $\frac{{}^{30}C_r}{{}^{30}C_{r-1}}=\frac{31-r}r$, so each term is $r(31-r)\,{}^{30}C_r$. $\sum r\,{}^{30}C_r=30\cdot2^{29}$ and $\sum r^2\,{}^{30}C_r=30\cdot31\cdot2^{28}$. The sum is $31\cdot30\cdot2^{29}-930\cdot2^{28}=(930-465)\cdot2^{29}=465\cdot2^{29}$. Answer: 465.

Q. If $\sum_{r=1}^{30}\frac{r^2\left({}^{30}C_r\right)^2}{{}^{30}C_{r-1}}=\alpha\times2^{29}$, then $\alpha$ is equal to _______

Explanation

Reviewed by Dr. Abubacker

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