Check sibling questions

Ex 9.4, 8 - Find sum of series whose nth term is n(n+1)(n+4) - Finding sum from nth number

Ex 9.4, 8 - Chapter 9 Class 11 Sequences and Series - Part 2

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Ex 9.4, 8 Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4). Given an = n(n + 1) (n + 4) = (n2 + n)(n + 4) = n2 (n + 4) + n(n + 4) = n3 + 4n2 + n2 + 4n = n3 + n2 + 4n2 + 4n = n3 + 5n2 + 4n The sum of n terms is = (n(n + 1)/( 2))^2+ 5 ((n(n + 1)(2n + 1))/( 6)) + 4((n(n + 1))/( 2)) = n2(n + 1)2/4 + (5(n(n + 1)(2n + 1)))/( 6) + 2(n(n + 1)) = n(n + 1)(n(n + 1)/4 " + " ((5(2n + 1)) )/( 6) "+ 2" ) = n(n + 1) (((3n(n + 1)) + (10(2n + 1)) +24)/( 12)) = n(n + 1) ((3n2 + 3n + 20n + 10 + 24 )/( 12)) = (𝑛(𝑛 + 1))/12 (3n2 + 23n + 34) Thus, the required sum is (𝑛(𝑛 + 1))/12 (3n2 + 23n + 34)

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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.