The sum [1+(1^3+2^3)/(1+2)+(1^3+2^3+3^3)/(1+2+3)++(1^3+2^3++15^3)/(1+2++15) - 12(1+2++15)]…

Question

The sum
$$1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+\cdots+\frac{1^3+2^3+\cdots+15^3}{1+2+\cdots+15} - \frac12(1+2+\cdots+15)$$
is equal to:

Accepted Answer

$T_n = \dfrac{\left(\frac{n(n+1)}{2}\right)^2}{\frac{n(n+1)}{2}} = \dfrac{n(n+1)}{2}$.

$$\sum_{n=1}^{15}T_n = \frac12\left[\frac{15\times16\times31}{6}+\frac{15\times16}{2}\right] = 680.$$

$$680 - \frac12\cdot\frac{15\times16}{2} = 680-60 = 620.$$

Solutions · 1