The sum of the series 12^2+24^2+36^2+ up to 10 terms is:

Question

The sum of the series $1\cdot2^2+2\cdot4^2+3\cdot6^2+\cdots$ up to 10 terms is:

Accepted Answer

The $n$-th term is $n(2n)^2=4n^3$.

$$S_{10}=4\sum_{n=1}^{10}n^3=4\left(\frac{10\times11}{2}\right)^2=4(55)^2=4\times3025=12100.$$

Solutions · 1