If _k=1^n a_k = 6n^3 for every n N, then _k=1^6(a_k+1-a_k36)^2 is equal to ________.

Question

If $\displaystyle\sum_{k=1}^{n} a_k = 6n^3$ for every $n \in \mathbb{N}$, then $\displaystyle\sum_{k=1}^{6}\left(\frac{a_{k+1}-a_k}{36}\right)^2$ is equal to \_\_\_\_\_\_\_\_.

Answer: 91

Accepted Answer

From $\sum_{k=1}^n a_k = 6n^3$, $a_n = 6n^3 - 6(n-1)^3 = 6\big[n^3-(n-1)^3\big] = 18n^2-18n+6$.

$$a_{k+1}-a_k = 18(k+1)^2 - 18(k+1) - \left(18k^2-18k\right) = 18(2k+1) - 18 = 36k.$$

So $\dfrac{a_{k+1}-a_k}{36} = k$, and
$$\sum_{k=1}^{6} k^2 = 1+4+9+16+25+36 = 91.$$

Solutions · 1