The sum of $n$ terms of the series whose $n^{th}$ term is $n(n + 1)$ is equal to.
- $\dfrac {n(n + 1)(n + 2)}{3}$
- $\dfrac{(n + 1)(n + 2)(n+3)}{12}$
- $n^2 (n + 2)$
- $n(n + 1)(n + 2)$
Reveal answer
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A
Correct answer
Explanation
The sum of n terms where the n-th term is n(n+1) = n^2 + n is the sum of the first n squares plus the sum of the first n integers. This equals [n(n+1)(2n+1)/6] + [n(n+1)/2], which simplifies to n(n+1)(n+2)/3.
AI explanation
The nth term of the series is T_n = n(n + 1), so the sum of n terms is the summation of (n^2 + n). Using the standard summation formulas, the sum of n^2 is n(n + 1)(2n + 1)/6 and the sum of n is n(n + 1)/2. Adding these gives a common denominator of 6 to produce [2n^3 + 3n^2 + n + 3n^2 + 3n] / 6, which simplifies to (2n^3 + 6n^2 + 4n) / 6. Factoring out 2n leaves 2n(n^2 + 3n + 2) / 6, and dividing by 2 results in n(n + 1)(n + 2) / 3.