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NCERT Exemplar · Q11

Q.Find the sum of the series (33−23)+(53−43)+(73−63)+…(3^3 - 2^3) + (5^3 - 4^3) + (7^3 - 6^3) + \ldots to

(i) nn terms
(ii) 1010 terms.
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Sum to nn terms is Sn=n(4n2+9n+6)S_n = n(4n^2 + 9n + 6); sum to 1010 terms is 49604960.

The general term is the difference of consecutive odd/even cubes:

tn=(2n+1)3−(2n)3.t_n = (2n+1)^3 - (2n)^3.

Expanding with (a+1)3−a3=3a2+3a+1(a+1)^3 - a^3 = 3a^2 + 3a + 1 where a=2na = 2n:

tn=3(2n)2+3(2n)+1=12n2+6n+1.t_n = 3(2n)^2 + 3(2n) + 1 = 12n^2 + 6n + 1.

(i) Sum to nn terms

Sn=∑k=1n(12k2+6k+1)=12⋅n(n+1)(2n+1)6+6⋅n(n+1)2+n.S_n = \sum_{k=1}^{n}(12k^2 + 6k + 1) = 12\cdot\frac{n(n+1)(2n+1)}{6} + 6\cdot\frac{n(n+1)}{2} + n. …

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