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Ex 9.4, 2 - Find sum 1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 + - Finding sum from nth number

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

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Transcript

Ex 9.4, 2 Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + … 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + … Step 1: Find nth term (an) Here, an = n ( n + 1) (n + 2) = (n2 + n) (n + 2) = n2(n + 2) + n(n + 2) = n3 + 2n2 + n2 + 2n = n3 + 3n2 + 2n Step 2: Finding sum of n terms = ((n(n+1))/2)^2+ 3 (n(n+1)(2n+1))/6 +2 (n(n+1))/2 = n2(n+1)2/4 + (n(n+1)(2n+1))/2 + n(n + 1) = (n(n+1))/2 ((n(n+1))/2 " + (2n + 1) + 2" ) = (n(n+1))/2 ((n2 + n)/2 " + 2n + 1 + 2" ) = (n(n+1))/2 ( (n2 + n)/2 + 2n + 3) = (n(n+1))/2 ((n2 + n + 4n + 6)/2) = (n(n+1))/2 ((n2 + 5n + 6)/2) = (n(n+1))/(2 × 2) (n2 + 5n + 6) = (n(n+1))/4 (n2 + 2n + 3n + 6) = n(n+1)/4 [n(n + 3) + 2(n + 3)] = (n(n+1)(n+2)(n+3))/4 Thus, the required sum is (n(n+1)(n+2)(n+3))/4

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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.