Having a problem with simplifying a sum given that $(k+1)^3-(k-1)^3=6 k^2+2$

Question

I was able to prove this first part (given in the title) however I am having trouble continuing with the second part of the question:

Any help or suggestions for how to solve this would be greatly appreciated. :)

Answer 1

We can use the telescopic sum. $$\sum_{k=1}^n(6k^2+2)=\sum_{k=1}^n((k+1)^3-(k-1)^3)=$$ $$=\sum_{k=1}^n((k+1)^3-k^3+(k^3-(k-1)^3))=(n+1)^3-1+(n^3-0).$$ Can you end it now?

Answer 2

Hint

Note that $$\sum_1^n 6k^2+2{=\sum_{k=1}^{n}(k+1)^3-(k-1)^3\\=\sum_{k=1}^{n}(k+1)^3-k^3+k^3-(k-1)^3\\=\sum_{k=1}^{n}(k+1)^3-k^3\\+\sum_{k=1}^{n}k^3-(k-1)^3}$$Now use the telescopic rule.