Question

Find the sum of the series $(3^3-2^3)+(5^3-4^3)+(7^3-6^3)+......$ to $10$ terms.

Answer 1

$(3^3-2^3)+(5^3-4^3)+...+({21}^3-{20}^3)$

$=(3-2)(3^2+6+2^2)+(5-4)(5^2+20+4^2)+...+(21-20)({21}^2+420+{20}^2)$

$=2^2+3^2+...+{21}^2+(2^2+2)+(4^2+4)+..+({20}^2+20)$

$=\frac{21\times 22\times 43}{6}-1+4(1+2^2+3^2+...+{10}^2)+2(1+2+...+10)$

$=3311-1+4\left(\frac{10\times 11\times 21}{6}\right)+\left(10\right)\left(11\right)$

$=3311-1+1540+110$

$=4960$