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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 103 of 106
Marks: +1, -0
Find the sum of the following series up to n terms :
131+13+231+3+13+23+331+3+5\frac{1^3}{1} + \frac{1^3+2^3}{1+3} + \frac{1^3+2^3+3^3}{1+3+5} + ...
Solution:  
We have 131+13+231+3+13+23+331+3+5\frac{1^3}{1} + \frac{1^3+2^3}{1+3} + \frac{1^3+2^3+3^3}{1+3+5} + ...
∴ n th term is
ana_n = 13+23+⋯+n31+3+5+⋯+→n terms\frac{1^3+2^3+\dots+n^3}{1+3+5+\dots+\to n\text{ terms}}
ana_n = [n(n+1)2]2n2[2+2(n−1)]\frac{\left[\frac{n(n+1)}{2}\right]^2}{\frac{n}{2}\left[2+2(n-1)\right]} = [n(n+1)2]2n(n)\frac{\left[\frac{n(n+1)}{2}\right]^2}{n(n)} = n2(n+1)2n2×4\frac{n^2 (n+1)^2}{n^2 \times 4}
= (n+1)24\frac{(n+1)^2}{4} = n2+2n+14\frac{n^2+2n+1}{4}
Hence, the sum to n terms is
SnS_n = ∑nk=1ak\sum\limits^{\underset{k=1}{n}} a_k = ∑nk=1k2+2k+14\sum\limits^{\underset{k=1}{n}} \frac{k^2+2k+1}{4} = 14[∑nk=1k2+2∑nk=1k+∑nk=11]\frac{1}{4} \left[ \sum\limits^{\underset{k=1}{n}} k^2 + 2 \sum\limits^{\underset{k=1}{n}} k + \sum\limits^{\underset{k=1}{n}} 1 \right]
=14[n(n+1)(2n+1)6+2n(n+1)2+n]\frac{1}{4} \left[ \frac{n(n+1)(2n+1)}{6} + \frac{2n(n+1)}{2} + n \right]
=n4[(n+1)(2n+1)6+(n+1)+1]= \frac{n}{4} \left[ \frac{(n+1)(2n+1)}{6} + (n+1) + 1 \right]
=n4[2n2+n+2n+1+6n+6+66]= \frac{n}{4} \left[ \frac{2n^2+n+2n+1+6n+6+6}{6} \right]
= n24(2n2+9n+13)\frac{n}{24} (2n^2+9n+13)
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