If _k=1^3131k31k-1-_k=1^3030k30k-1=(60!)(30!)(31!) where R, then the value of 16 is equal …

Question

If $\displaystyle\sum_{k=1}^{31}\binom{31}{k}\binom{31}{k-1}-\sum_{k=1}^{30}\binom{30}{k}\binom{30}{k-1}=\dfrac{\alpha(60!)}{(30!)(31!)}$ where $\alpha\in\mathbb R$, then the value of $16\alpha$ is equal to

Accepted Answer

Using Vandermonde-type identities: $\displaystyle\sum_{k=1}^{31}\binom{31}{k}\binom{31}{k-1}={}^{62}C_{30}$, and $\displaystyle\sum_{k=1}^{30}\binom{30}{k}\binom{30}{k-1}={}^{60}C_{29}$.

$${}^{62}C_{30}-{}^{60}C_{29}=\frac{62\cdot61\cdot60!}{30!\,32!}-\frac{60!}{29!\,31!}=\frac{60!}{30!\,31!}\left[\frac{62\times61}{32}-30\right]=\frac{\alpha\cdot60!}{30!\,31!}$$

$$\alpha=\frac{62\times61}{32}-30\quad\Rightarrow\quad16\alpha=\frac{62\times61-30\times32}{2}=\frac{3782-960}{2}=\frac{2822}{2}=1411$$

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