The sum . Up to 10^th bracket is

JEE MainMathematicsModerateSequences and Series

The sum
$\left( 1^{2} \right) + \left( 1^{2} + 2^{2} \right) + \left( 1^{2} + 2^{2} + 3^{2} \right)$ .
Up to 10^th bracket is

Show answer and solution

T_{n} = 1^{2} + 2^{2} + 3^{2} + \ldots\ldots.. + n^{2}

T_{n} = \frac{n(n + 1)(2n + 1)}{6} = \frac{2n^{3} + 3n^{2} + n}{6}

S_{n} = \sum T_{n} = \frac{1}{6}\lbrack 2\Sigma n^{3} + 3\Sigma n^{2} + \Sigma n\rbrack

= \frac{1}{6}\{ 2{\lbrack\frac{n(n + 1)}{2}\rbrack}^{2} + \frac{3n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}\}

S_{n} = \frac{n(n + 1)^{2}(n + 2)}{12}

Source: Grass-G / M-16 Mathematics Paper 25 Nov First.docx

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