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Ex 9.4, 3 - Find sum 3 x 12 + 5 x 22 + 7 x 32 + .. - Chapter 9 - Ex 9.4

Ex 9.4, 3 - Chapter 9 Class 11 Sequences and Series - Part 2
Ex 9.4, 3 - Chapter 9 Class 11 Sequences and Series - Part 3

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Transcript

Ex 9.4, 3 Find the sum to n terms of the series 3 12 + 5 22 + 7 32 + Hence an = (nth term of 3, 5, 7, ) (nth term of 12,22,32,..) an = (2n + 1)n2 an = 2n3 + n2 Now, finding sum of n terms of series = 2((n(n+1))/2)^2 + (n(n+1)(2n+1))/6 = (n(n+1))/2 ("n(n + 1) + " (2n+1)/3) = (n(n+1))/2 ("n2 + n + " (2n+1)/3) = (n(n+1))/2 ("n2 + n + " (2n+1)/3) = (n(n+1))/2 ((3n2 + 3n + 2n + 1)/3) = (n(n+1))/2 ((3n2 + 5n + 1)/3) = (n(n + 1)(3n2 + 5n + 1))/6 Thus, the required sum is (n(n + 1)(3n2 + 5n + 1))/6

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.