If T_n = 1n(1^2+2^2++n^2), the sum T_1+T_2++T_10 (up to 10 terms) is equal to:

Question

If $T_n = \dfrac1n\left(1^2+2^2+\cdots+n^2\right)$, the sum $T_1+T_2+\cdots+T_{10}$ (up to 10 terms) is equal to:

Accepted Answer

$$T_n = \frac{n(n+1)(2n+1)}{6n} = \frac16(2n^2+3n+1).$$

$$\sum_{n=1}^{10}T_n = \frac16\left[2\sum n^2+3\sum n+\sum 1\right] = \frac16\left[2(385)+3(55)+10\right] = \frac{10}{6}\left[77+\frac{33}{2}+1\right] = \frac{315}{2}.$$

Solutions · 1