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AP EAMCET · Maths · Sequences and Series

\(1^2+\left(1^2+2^2\right)+\left(1^2+2^2+3^2\right)+\ldots+\)
\(\left(1^2+2^2+\ldots+n^2\right)=\)

  1. A \(\frac{n(n+1)(n+2)}{12}\)
  2. B \(\frac{n(n+1)(2 n+1)}{6}\)
  3. C \(\frac{n(n+1)^2(n+2)}{12}\)
  4. D \(\frac{n(n+1)(n+2)(n+3)}{12}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{n(n+1)^2(n+2)}{12}\)

Step-by-step Solution

Detailed explanation

\begin{aligned} & \text { } 1^2+\left(1^2+2^2\right)+\left(1^2+2^2+3^2\right)+\ldots \\ & +\left(1^2+2^2+\ldots+n^2\right) \\ & =\sum_{i=1}^n\left(1^2+2^2+3^2+\ldots+i^2\right)=\sum_{i=1}^n \sum_{j=1}^i j^2 \\ & =\sum_{i=1}^n \frac{i(i+1)(2 i+1)}{6}\left\{\because \Sigma…

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