1.2.4 + 2.3.7 +3.4.10 + ...

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Solution

Let $T_{n}$ be the nth term of the given series.

Thus, we have:

$T_{n} = n (n + 1) (3 n + 1) = n (3 n^{2} + 4 n + 1) = (3 n^{3} + 4 n^{2} + n)$

Now, let $S_{n}$ be the sum of n terms of the given series.

Thus, we have:
$S_{n} = \sum_{k = 1}^{n} T_{k}$

$\Rightarrow S_{n} = {\sum 3}_{k = 1}^{n} k^{3} + {\sum 4}_{k = 1}^{n} k^{2} + \sum_{k = 1}^{n} k \Rightarrow S_{n} = 3 \sum_{k = 1}^{n} k^{3} + {4 \sum}_{k = 1}^{n} k^{2} + \sum_{k = 1}^{n} k \Rightarrow S_{n} = \frac{3 n^{2} {(n + 1)}^{2}}{4} + \frac{4 n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2} \Rightarrow S_{n} = \frac{3 n^{2} {(n + 1)}^{2}}{4} + \frac{2 n (n + 1) (2 n + 1)}{3} + \frac{n (n + 1)}{2} \Rightarrow S_{n} = \frac{n (n + 1)}{2} (\frac{3 n (n + 1)}{2} + \frac{4 (2 n + 1)}{3} + 1) \Rightarrow S_{n} = \frac{n (n + 1)}{2} (\frac{3 n^{2} + 3 n}{2} + \frac{8 n + 4}{3} + 1) \Rightarrow S_{n} = \frac{n (n + 1)}{2} (\frac{9 n^{2} + 9 n + 16 n + 8 + 6}{6}) \Rightarrow S_{n} = \frac{n (n + 1)}{12} (9 n^{2} + 25 n + 14)$

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Similar questions

The sum of series 1.2.4 + 2.3.5 + 3.4.6 + ----------- 20 terms.

\frac{3}{1.2.4} + \frac{4}{2.3.5} + \frac{5}{3.4.6} + . . . .

n t h

n

1.2.4 + 2.3.5 + 3.4.6 + . . . .

s_{n} = \frac{3}{1.2.4} + \frac{4}{2.3.5} + \frac{5}{3.4.6} + \dots \dots + \frac{n + 2}{n (n + 1) (n + 3)}

l i m n \to \infty s_{n}

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