Find the 20^th term and the sum of 20 terms of the series 2 × 4 + 4 × 6 + 6 × 8 + ...

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Solution

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

Thus, we have:

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

For n = 20, we have:

$T_{20} = 4 n^{2} + 4 n = 4 {(20)}^{2} + 4 (20) = 1600 + 80 = 1680$

Therefore, the 20th term of the given series is 1680.

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

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

$\Rightarrow S_{n} = \sum_{k = 1}^{n} (4 k^{2} + 4 k) \Rightarrow S_{n} = {4 \sum}_{k = 1}^{n} k^{2} + 4 \sum_{k = 1}^{n} k$

For n = 20, we have:

$S_{20} = {4 \sum}_{k = 1}^{20} k^{2} + 4 \sum_{k = 1}^{20} k \Rightarrow S_{20} = \frac{4 (20) (21) (41)}{6} + \frac{4 (20) (21)}{2} \Rightarrow S_{20} = (40) (7) (41) + (40) (21) \Rightarrow S_{20} = 11480 + 840 = 12320$

Hence, the sum of the first 20 terms of the series is 12320.

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Similar questions

Find the 20 $^{t h}$ term and the sum of 20 terms of the series :

$2 \times 4 + 4 \times 6 + 6 \times 8 + . . . .$

Find the 20^th term of the series 2 × 4 + 4 × 6 + 6 × 8 + … + n terms.

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2 \times 4 + 4 \times 6 + 6 \times 8 + . . . + n

20^{t h}

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