JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(S_{ k }=\frac{1+2+\ldots .+ K }{ K }\) and \(\sum_{j=1}^n S_j^2=\frac{n}{A}\left( Bn ^2+ Cn + D \right)\), where \(A , B , C , D \in N\) and \(A\) has least value. Then
- A \(A + B\) is divisible by \(D\)
- B \(A+B=5(D-C)\)
- C \(A + C + D\) is not divisible by \(B\)
- D \(A + B + C + D\) is divisible by \(5\)
Answer & Solution
Correct Answer
(A) \(A + B\) is divisible by \(D\)
Step-by-step Solution
Detailed explanation
\(S _{ k }=\frac{ k +1}{2}\) \(S _{ k }^2=\frac{ k ^2+1+2 k }{4}\) \(\therefore \sum \limits_{ j -1}^{ n } S _{ j }^2=\frac{1}{4}\left[\frac{ n ( n +1)(2 n +1)}{6}+ n + n ( n +1)\right]\) \(=\frac{ n }{4}\left[\frac{( n +1)(2 n +1)}{6}+1+ n +1\right]\)…
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