Question

Let S_K = \(\frac{1+2+...+ K}{K}\) and \(\displaystyle\sum_{j=1}^{n}S_j^2=\frac{n}{A}(Bn^2+Cn+D)\), where A,B,C,D∈N and A has least value. Then

• A. A+B+C+D is divisible by 5
• B. A+B is divisible by D
• C. A+B=5(D-C)
• D. A+C+D is not divisible by B

Hint:

Use algebraic manipulation to rewrite expressions and compare coefficients systematically.

Answer 1

The Correct Option is B

We have: \[ S_K = \frac{k+1}{2}, \] \[ S_k^2 = \frac{k^2 + 1 + 2k}{4}. \] Thus, \[ \sum_{j=1}^{n} S_j^2 = \frac{1}{4} \sum_{j=1}^{n} (n(n+1)) + \sum_{j=1}^{n} (n(n+1)). \] We can simplify this further: \[ S_j = \frac{(n+1)(2n+1)}{6}. \] Solving for the other terms: \[ n = \frac{(n+1)}{6}. \] From this we can derive \(A = 24\), \(B = 2\), \(C = 9\), and \(D = 13\).
\[ A + B \text{ is divisible by } D \]