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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 74 of 106
Marks: +1, -0
12+(12+22)+(12+22+32)1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + .....
Solution:  
In the given series
ana_n = 1+221^{+}2^2 + ... + n2n^2 = n(n+1)(2n+1)6\frac{n(n+1)(2n+1)}{6} = (n2+n)(2n+1)6\frac{(n^2+n)(2n+1)}{6}
= 3n3+2n2+n2+n6\frac{3n^3+2n^2+n^2+n}{6} = 2n3+3n2+n6\frac{2n^3+3n^2+n}{6}
Hence, the sum to n terms is
SnS_n = ∑nk=1ak\sum\limits^{\underset{k=1}{n}} a_k = 16∑kk=1(2k3+3k2+k)\frac{1}{6} \sum\limits_{\underset{k=1}{k}} (2k^3+3k^2+k) = 16[2∑nk=1k3+3∑nk=1k2+∑nk=1k]\frac{1}{6}\left[2\sum\limits^{\underset{k=1}{n}} k^3 + 3\sum\limits^{\underset{k=1}{n}} k^2 + \sum\limits^{\underset{k=1}{n}} k\right]
=
16[2(n(n+1)2)2+3(n(n+1)(2n+1)6)+n(n+1)2]\frac{1}{6}\left[2\left(\frac{n(n+1)}{2}\right)^2+3\left(\frac{n(n+1)(2n+1)}{6}\right)+\frac{n(n+1)}{2}\right]
= n(n+1)2×6\frac{n(n+1)}{2 \times 6} [n (n + 1) + (2n + 1) + 1] = n(n+1)12[n2+n+2n+2]\frac{n(n+1)}{12}[n^2+n+2n+2]
= n(n+1)12\frac{n(n+1)}{12} (n + 1) (n + 2) = n(n+1)2(n+2)12\frac{n(n+1)^2(n+2)}{12}
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