Chapter 10: Problem 56
The sum of first \(n\) terms of the series \(1^{2}+2.2^{2}+3^{2}+2.4^{2}+5^{2}+5.6^{2}+\ldots\) is \(\frac{n(n+1)^{2}}{2}\) when \(n\) is even. When \(n\) is odd, the sum is (A) \(\frac{n^{2}(n+1)}{2}\) (B) \(\frac{n(n+1)^{2}}{2}\) (C) \(\left[\frac{n(n+1)}{2}\right]^{2}\) (D) \(\frac{n(n+1)}{2}\)
Short Answer
Step by step solution
Understanding the Series
Calculate for Even n
Pattern Matching for Odd n
Examine the Odd Pattern
Relate to the Sum for Even Terms
Choose the Correct Option for Odd n
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