The sum of the series $$3$$×$$1312+513+2312+22+713+23+3312+22+32+$$…$$+$$ upto to $$10th$$ term is

$$Tn=2n+1nn+122nn+12n+16=32n2+nS10=$$∑$$T10=32$$∑$$110n2+$$∑$$110n=3210$$×$$11$$×$$216+10$$×$$112=660$$

Series of natural numbers

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$The sum of the series \frac{3 \times 1^{3}}{1^{2}} + \frac{5 (1^{3} + 2^{3})}{1^{2} + 2^{2}} + \frac{7 (1^{3} + 2^{3} + 3^{3})}{1^{2} + 2^{2} + 3^{2}} + \dots + upto to 10^{th} term is$

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Solution

$\begin{array}{l} T_{n} = \frac{(2 n + 1) {[\frac{n (n + 1)}{2}]}^{2}}{\frac{n (n + 1) (2 n + 1)}{6}} = \frac{3}{2} (n^{2} + n) \\ S_{10} = \sum T_{10} = \frac{3}{2} (\sum_{1}^{10} n^{2} + \sum_{1}^{10} n) = \frac{3}{2} [\frac{10 \times 11 \times 21}{6} + \frac{10 \times 11}{2}] \\ = 660 \end{array}$

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Series of natural numbers

Remember concepts with our Masterclasses.

The sum of the series 3×1312+513+2312+22+713+23+3312+22+32+…+ upto to 10th term is

Tn=2n+1nn+122nn+12n+16=32n2+nS10=∑T10=32∑110n2+∑110n=3210×11×216+10×112=660

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$The sum of the series \frac{3 \times 1^{3}}{1^{2}} + \frac{5 (1^{3} + 2^{3})}{1^{2} + 2^{2}} + \frac{7 (1^{3} + 2^{3} + 3^{3})}{1^{2} + 2^{2} + 3^{2}} + \dots + upto to 10^{th} term is$