Check sibling questions

Ex 9.4, 5 - Find sum of series 52 + 62 + 72 + ... + 202 - Ex 9.4

Ex 9.4, 5 - Chapter 9 Class 11 Sequences and Series - Part 2
Ex 9.4, 5 - Chapter 9 Class 11 Sequences and Series - Part 3


Transcript

Ex 9.4, 5 Find the sum to n terms of the series 52 + 62 + 72 + .. + 202 52 + 62 + 72 + .. + 202 = (12 + 22 + 32 + 42 + 52 + 62 + + 202) (12 + 22 + 32 + 42) We know that Sum of square of n natural number is i.e. (12 + 22 + + n2) = (n(n+1)(2n+1))/6 For 12 + 22 + + 202 n = 20 Putting n = 20 in (2) 12 + 22 + + 202 = (20(20 + 1)(2(20) + 1))/6 = (20(21)(40 + 1))/6 = (20 (21) (41))/6 = (20 21 41)/6 = 10 7 41 = 2870 Thus, 12 + 22 + + 202 = 2870 For 12 + 22 + 32 + 42 n = 4 Putting n = 4 in (12 + 22 + + n2) = ( ( +1)(2 +1))/6 12 + 22 + 32 + 42 = (4(4 + 1)(2(4) + 1))/6 = (4 5 9)/6 = 2 5 3 = 30 Now, from (1) 52 + 62 + 72 + .. + 202 = (12 + 22 + 32 + 42 + 52 + 62 + + 202) (12 + 22 + 32 + 42) Putting values = 2870 30 = 2840 Thus, the required sum is 2840

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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.