Let S_k = 1+2++kk and _j=1^nS_j^2 = nA(Bn^2+Cn+D), where A,B,C,DN and A has the least poss…

Question

Let $S_k = \dfrac{1+2+\cdots+k}{k}$ and $\displaystyle\sum_{j=1}^{n}S_j^2 = \dfrac{n}{A}\big(Bn^2+Cn+D\big)$, where $A,B,C,D\in\mathbb{N}$ and $A$ has the least possible value. Then:

Accepted Answer

$S_k = \dfrac{k+1}{2}$, so $S_k^2 = \dfrac{k^2+2k+1}{4}$.

$$\sum_{j=1}^{n}S_j^2 = \frac14\left[\frac{n(n+1)(2n+1)}{6}+n(n+1)+n\right] = \frac{n}{24}\big(2n^2+9n+13\big).$$

So $A=24, B=2, C=9, D=13$. Checking: $A+B=26$, and $26$ is divisible by $D=13$ --- Option A holds.

Solutions · 1