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The value of $$\frac{1}{2} + \frac{1^2+2^2}{6} + \frac{1^2+2^2+3^2}{12} + \frac{1^2+2^2+3^2+4^2}{20} + \cdots + \frac{1^2+2^2+\cdots+60^2}{3660}$$ is
Correct Answer: 620
The $$n$$th term has numerator $$\frac{n(n+1)(2n+1)}{6}$$ and denominator $$n(n+1)$$, so it simplifies to $$\frac{2n+1}{6}$$. Summing for $$n = 1$$ to 60 gives $$\frac{1}{6}\left(2 \times \frac{60 \times 61}{2} + 60\right) = \frac{3660+60}{6}$$. This equals $$\frac{3720}{6} = 620$$.
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