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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 104 of 106
Marks: +1, -0
Show that
1×22+2×32+…+n×(n+1)212×2+22×3+…+n2×(n+1)\frac{1\times 2^2+2\times 3^2+\ldots+n\times (n+1)^2}{1^2\times 2+2^2\times 3+\ldots+n^2\times (n+1)}
= 3n+53n+1\frac{3n+5}{3n+1}
Solution:  
We have
1×22+2×32+…+n×(n+1)212×2+22×3+…+n2×(n+1)\frac{1\times 2^2+2\times 3^2+\ldots+n\times (n+1)^2}{1^2\times 2+2^2\times 3+\ldots+n^2\times (n+1)}
∴ nth term is ana_n = n(n+1)2n2(n+1)\frac{n(n+1)^2}{n^2(n+1)} = n3+2n2+nn2+n3\frac{n^3+2n^2+n}{n^2+n^3}
Hence, the sum to n terms is SnS_n = ∑k=1nak\sum\limits_{k=1}^{n} a_k = ∑k=1nk3+2k2+kk3+k2\sum\limits_{k=1}^{n} \frac{k^3+2k^2+k}{k^3+k^2}
= ∑k=1nk3+2∑k=1nk2+∑k=1nk∑k=1nk3+∑k=1nk2\frac{\sum\limits_{k=1}^{n} k^3 + 2\sum\limits_{k=1}^{n} k^2 + \sum\limits_{k=1}^{n} k}{\sum\limits_{k=1}^{n} k^3 + \sum\limits_{k=1}^{n} k^2} =
[n(n+1)2]2+2n(n+1)(2n+1)6+n(n+1)2[n(n+1)2]2+n(n+1)(2n+1)6\frac{\left[\frac{n(n+1)}{2}\right]^2 + \frac{2n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}}{\left[\frac{n(n+1)}{2}\right]^2 + \frac{n(n+1)(2n+1)}{6}}
=
n(n+1)2[n(n+1)2+2(2n+1)3+1]n(n+1)2[n(n+1)2+2n+13]\frac{\frac{n(n+1)}{2} \left[ \frac{n(n+1)}{2} + \frac{2(2n+1)}{3} + 1 \right]}{\frac{n(n+1)}{2} \left[ \frac{n(n+1)}{2} + \frac{2n+1}{3} \right]}
= 3n2+3n+8n+4+63n2+3n+4n+2\frac{3n^2+3n+8n+4+6}{3n^2+3n+4n+2} = 3n2+11n+103n2+7n+2\frac{3n^2+11n+10}{3n^2+7n+2}
= (3n+5)(n+2)(3n+1)(n+2)\frac{(3n+5)(n+2)}{(3n+1)(n+2)} = 3n+53n+1\frac{3n+5}{3n+1}
Hence,
1×22+2×32+…+n×(n+1)212×2+22×3+…+n2×(n+1)\frac{1\times 2^2+2\times 3^2+\ldots+n\times (n+1)^2}{1^2\times 2+2^2\times 3+\ldots+n^2\times (n+1)}
= 3n+53n+1\frac{3n+5}{3n+1} = 3n+53n+1\frac{3n+5}{3n+1}
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