Finding sum from nth number

Chapter 8 Class 11 Sequences and Series
Concept wise

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### Transcript

Question 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)

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.