1²+(1²+2²) + (1²+2²+3²) + ... + (1²+2²++n²) =

Correct Answer is : n(n+1)2(n+2)12n(n+1)²(n+2)/1212n(n+1)2(n+2)

Correct Answer is : n(n+1)2(n+2)12n(n+1)²(n+2)/1212n(n+1)2(n+2)

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AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 3 of 160

Marks: +1, -0

1^2+(1^2+2^2)

(1^2+2^2+3^2)

(1^2+2^2+\dots+n^2)

Solution:

Let

S=[1^{2}+(1^{2}+2^{2})+(1^{2}+2^{2}+3^{2})+\dots+(1^{2}+2^{2}+\dots+n^{2})]

S \;\;=\sum\limits_{i=1}^{n} \sum\limits_{j=1}^{i} j^{2}

\;\;=\sum\limits_{i=1}^{n} \;\frac{2 i^{3}+3 i^{2}+i}{6}

=\;\frac{1}{3}\left[\frac{n(n+1)}{2}\right]^{2}+\;\frac{1}{2}\left[\frac{n(n+1)(2 n+1)}{6}\right]+\;\frac{1}{6}\left[\frac{n(n+1)}{2}\right]

Simplify further.

S \;\;=\;\frac{1}{3} \;\frac{n^{2}(n+1)^{2}}{4}+\;\frac{1}{2} \;\frac{n(n+1)(2 n+1)}{6}+\;\frac{1}{6} \;\frac{n(n+1)}{2}

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

\;\;=\;\frac{n(n+1)}{12}[(n+1)(n+2)]

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

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