JEE Mains · Maths · STD 11 - 8. sequence and series
If \(\displaystyle\sum_{k=1}^{n} a_k = 6n^3\), then \(\displaystyle\sum_{k=1}^{6} \left(\dfrac{a_{k+1} - a_k}{36}\right)^2\) is equal to _______.
- A 90
- B 92
- C 91
- D 94
Answer & Solution
Correct Answer
(C) 91
Step-by-step Solution
Detailed explanation
Given \(\displaystyle\sum_{k=1}^{n} a_k = 6n^3\). Let \(S_n = 6n^3\). The \(n\)-th term of the sequence is given by \(a_n = S_n - S_{n-1}\). \(a_n = 6n^3 - 6(n-1)^3\) \(a_n = 6[n^3 - (n^3 - 3n^2 + 3n - 1)]\) \(a_n = 6(3n^2 - 3n + 1)\) Replacing \(n\) with \(k\) and \(k+1\), we…
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