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Ta có:
$\sum_{i=1}^{n}i^3=\Big(\frac{n(n+1)}{2}\Big)^2$
$\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$
Ta có:
$\sum_{i=1}^{100}iC_{i+1}^2=\sum_{i=1}^{100}\frac{i^2(i+1)}{2}$
$=\frac{1}{2}\sum_{i=1}^{100}i^3+\frac{1}{2}\sum_{i=1}^{100}i^2$
$=\frac{1}{2}.\Big(\frac{100.101}{2}\Big)^2+\frac{1}{2}.\frac{100.101.201}{6}=12920425$
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