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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 78 of 106
Marks: +1, -0
(2n−1)2(2n-1)^2
Solution:  
We have
ana_n = (2n−1)2(2n-1)^2 = 4n2−4n+14n^2-4n+1
Hence, the sum of n terms is
SnS_n = 4[n(n+1)(2n+1)6]4\left[\frac{n(n+1)(2n+1)}{6}\right] - 4[n(n+1)2]4\left[\frac{n(n+1)}{2}\right] + n
= n(n+1)2[4(2n+1)3−4]\frac{n(n+1)}{2}\left[\frac{4(2n+1)}{3}-4\right] + n
= n(n+1)2[8n+4−123]+n\frac{n(n+1)}{2}\left[\frac{8n+4-12}{3}\right]+n
= n(n+1)(8n−8)6+n\frac{n(n+1)(8n-8)}{6}+n = n(n+1)(8n−8)+6n6\frac{n(n+1)(8n-8)+6n}{6} = n[8n2−8n+8n−8+6]6\frac{n\left[8n^2-8n+8n-8+6\right]}{6}
= n[8n2−2]6\frac{n\left[8n^2-2\right]}{6} = n3\frac{n}{3} = n/3n/3 (2n + 1) (2n - 1)
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