Progression (Sequence and Series)HardBloom L3
Question
The sum $2 \cdot 3 + 3 \cdot 6 + 4 \cdot 11 + 5 \cdot 18 + \ldots. +$ upto $n$ terms is equal to
Options
A.$\frac{n}{12}\left( 3n^{3} + 7n^{2} + 2\ln + 36 \right)$
B.$\frac{n}{12}\left( 3n^{3} - 10n^{2} + 22n - 36 \right)$
C.$\frac{n}{12}\left( 3n^{3} + 10n^{2} + 21n + 38 \right)$
D.$\frac{n}{12}\left( 3n^{3} + 8n^{2} + 20n + 37 \right)$
Solution
Sol. [Key Concept: General term is a polynomial in $n$; split into $\sum\ n^{3,}\sum\ n^{2,}\sum\ n$.]
$$S_{n} = \sum_{r = 1}^{n}\mspace{2mu}(r + 1)\left( r^{2} + 2 \right) = \frac{n^{2}(n + 1)^{2}}{4} + \frac{n(n + 1)(2n + 1)}{6} + n(n + 1) + 2n$$
$$= \frac{n}{12}\left( 3n^{3} + 10n^{2} + 2\ln + 38 \right)$$
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