AP EAMCET · Maths · Sequences and Series
For any integer \(n \geq 1\), the sum \(\sum_{k=1}^n k(k+2)\) is equal to
- A \(\frac{n(n+1)(n+2)}{6}\)
- B \(\frac{n(n+1)(2n+1)}{6}\)
- C \(\frac{n(n+1)(2n+7)}{6}\)
- D \(\frac{n(n+1)(2n+9)}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{n(n+1)(2n+7)}{6}\)
Step-by-step Solution
Detailed explanation
Now, \(\begin{aligned} & \sum_{k=1}^n k(k+2) \\ & =\sum_{k=1}^n\left(k^2+2 k\right)=\sum_{k=1}^n k^2+2 \sum_{k=1}^n k \\ & =\frac{n(n+1)(2 n+1)}{6}+\frac{2 \cdot n(n+1)}{2} \\ & =n(n+1)\left(\frac{2 n+1}{6}+1\right) \\ & =\frac{n(n+1)(2 n+7)}{6}\end{aligned}\)
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