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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 75 of 106
Marks: +1, -0
n (n + 1) (n + 4)
Solution:  
We have
ana_n = n(n + 1) (n + 4) = (n2n^2 + n) (n + 4) = n3+4n2+n2n^3 + 4n^2 + n^2 + 4n = n3+5n2n^3 + 5n^2 + 4n
Hence, the sum to n terms is
SnS_n = ∑k=1nak\sum\limits_{k=1}^{n} a_k = ∑k=1n(k3+5k2+4k)\sum\limits_{k=1}^{n} (k^3+5k^2+4k) = ∑k=1nk3+5∑k=1nk2+4∑k=1nk\sum\limits_{k=1}^{n} k^3+5\sum\limits_{k=1}^{n} k^2+4\sum\limits_{k=1}^{n} k
= [n(n+1)2]2\left[\frac{n(n+1)}{2}\right]^2 + 5[n(n+1)(2n+1)6]5\left[\frac{n(n+1)(2n+1)}{6}\right] + 4[n(n+1)2]4\left[\frac{n(n+1)}{2}\right]
=n(n+1)2[n(n+1)2+5(2n+1)3+4]\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{5(2n+1)}{3}+4\right]
=n(n+1)2[3n2+3n+20n+10+246]\frac{n(n+1)}{2}\left[\frac{3n^2+3n+20n+10+24}{6}\right]
= n(n+1)2[3n2+23n+346]\frac{n(n+1)}{2}\left[\frac{3n^2+23n+34}{6}\right] = n(n+1)(n+2)(3n+17)12\frac{n(n+1)(n+2)(3n+17)}{12}
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