Question

Find the sum of the series.
${11}^3+{12}^3+{13}^3+....+{28}^3$

Answer 1

Next we consider ${11}^3+{12}^3+....+{28}^3$
$=(1^3+2^3+3^3+...+{28}^3)-(1^3+2^3+...+{10}^3)$
$=\sum _1^{28}{n}^3-\sum _1^{10}{n}^3$
$={\left[\frac{28(28+1)}{2}\right]}^2-{\left[\frac{10(10+1)}{2}\right]}^2$
$={406}^2-{55}^2=(406+55)(406-55)$
$=\left(461\right)\left(351\right)=161811$.