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# 3 × 12 + 5 ×22 + 7 × 32 + ... - Mathematics

3 × 12 + 5 ×22 + 7 × 32 + ...

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#### Solution

Let $T_n$ be the nth term of the given series.
Thus, we have:

$T_n = \left( 2n + 1 \right) n^2 = 2 n^3 + n^2$

Now, let $S_n$ be the sum of n terms of the given series.
Thus, we have:

$S_n = \sum^n_{k = 1} T_k$

$\Rightarrow S_n = \sum^n_{k = 1} \left( 2 k^3 + k^2 \right)$

$\Rightarrow S_n = {2\sum}^n_{k = 1} k^3 + \sum^n_{k = 1} k^2$

$\Rightarrow S_n = \left[ \frac{2 n^2 \left( n + 1 \right)^2}{4} + \frac{n\left( n + 1 \right)\left( 2n + 1 \right)}{6} \right]$

$\Rightarrow S_n = \left[ \frac{n^2 \left( n + 1 \right)^2}{2} + \frac{n\left( n + 1 \right)\left( 2n + 1 \right)}{6} \right]$

$\Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left[ n\left( n + 1 \right) + \frac{2n + 1}{3} \right]$

$\Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left( \frac{3 n^2 + 3n + 2n + 1}{3} \right)$

$\Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left( \frac{3 n^2 + 5n + 1}{3} \right)$

$\Rightarrow S_n = \frac{n\left( n + 1 \right)}{6}\left( 3 n^2 + 5n + 1 \right)$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 21 Some special series
Exercise 21.1 | Q 7 | Page 10
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