Sequences & Series
Sum of Squares of Partial Sums
nta_pyq_2023_apr
Grade 11
Question:
Let $S_K=\frac{1+2+\cdots+K}{K}$ and $\displaystyle\sum_{j=1}^n S_j^2=\frac{An(Bn^2+Cn+D)}{?}$ with $A,B,C,D\in\mathbb{N}$, $A$ least. Then
$A+C+D$ not divisible by $D$
$A+B=5(D-C)$
$A+B+C+D$ divisible by 5
$A+B$ divisible by $D$
Step-by-Step Solution
Key Concept: $S_K=\frac{K+1}{2}$. $\sum S_j^2=\frac{1}{4}\sum(j+1)^2=\frac{n(2n^2+9n+13)}{24}$. So $A=24,B=2,C=9,D=13$.
$A=24,B=2,C=9,D=13$. Option (4).
Correct Answer: 4