Ex 9.4, 10 - Find sum of series, nth terms is (2n - 1)2 - Ex 9.4

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
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Ex 9.4, 10 Find the sum to n terms of the series whose nth terms is given by (2n – 1)2 Given an = (2n – 1)2 =(2n)2 + (1)2 – 2(2n)(1) = 4n2 + 1 – 4n = 4n2 – 4n + 1 Sum of n terms is = 4((n(n+1)(2n+1))/6) – 4 (n(n+1)/2) + n = n ("4" (n(n+1)(2n+1))/6 " – 4" (𝑛(𝑛 + 1))/2 " + 1" ) = n (2/3 " (n + 1)(2n + 1) – 2(n + 1) + 1" ) = n ((2(n + 1)(2n + 1) − 6(n + 1) + 3)/3) = 𝑛/3 [(2n + 2)(2n + 1) – 6n – 6 + 3 ] = 𝑛/3 [4n2 + 2n + 4n + 2 – 6n – 3] = 𝑛/3 [4n2 + 6n – 6n – 1] = 𝑛/3 [4n2 – 1] = 𝑛/3 [(2n)2 – (1)2] = 𝑛/3 [(2n – 1) (2n + 1)] Hence, the required sum is 𝑛/3 [(2n – 1 ) (2n + 1)]

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