For n≥2, let an=∑r=0n 1Cr2, then valueof bn=∑r=1n 1r2Cr2 equals

For $n \geq 2$ , let $a_{n} = \sum_{r = 0}^{n} \frac{1}{C_{r}^{2}}$ , then value
of $b_{n} = \sum_{r = 1}^{n} \frac{1}{r^{2} C_{r}^{2}}$ equals

For $r \geq 1, r C_{r} = r (^{n} C_{r}) = n (^{n - 1} C_{r - 1})$

Thus, $b_{n} = \sum_{r = 1}^{n} \frac{1}{n^{2} {(n - 1 C_{r})}^{2}} = \frac{1}{n^{2}} a_{n - 1}$

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