The sum of the series 1^2+22^2+3^2+24^2+5^2+26^2++2(2m)^2 is:

Question

The sum of the series $1^2+2\cdot2^2+3^2+2\cdot4^2+5^2+2\cdot6^2+\cdots+2(2m)^2$ is:

Accepted Answer

Group into $m$ pairs: pair $i$ is $(2i-1)^2+2(2i)^2=12i^2-4i+1$.

$$\sum_{i=1}^{m}(12i^2-4i+1)=2m(m+1)(2m+1)-2m(m+1)+m=m\big[2(m+1)(2m+1)-2(m+1)+1\big]=m\big[4m(m+1)+1\big]=m(2m+1)^2.$$

(Check $m=2$: series $=1+8+9+32=50=2\times25=2(5)^2$ .)

Solutions · 1