Ex 9.4, 7 - Find sum 12 + (12 + 22) + (12 + 22 + 32) + ... - Ex 9.4

Ex 9.4, 7 - Chapter 9 Class 11 Sequences and Series - Part 2
Ex 9.4, 7 - Chapter 9 Class 11 Sequences and Series - Part 3

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Transcript

Question 7 Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + … Step 1: Find nth term (an) Here an = (12 + 22 + 32 +…….+ n2) an = (n(n+1)(2n+1))/6 = ((n2 + n)(2n + 1))/6 = ((2n3 + n2 + 2n2 + n))/6 = ((2n3 + 3n2 + n))/6 Now, Sum of n terms is = 1/6 ("2" (((𝑛)(𝑛 + 1))/2)^2 " + 3" (((𝑛)(𝑛 +1)(2𝑛 +1))/6)" + " (𝑛(𝑛 + 1))/2) = 1/6 ("2" 𝑛2(𝑛 + 1)2/4 " + " (3𝑛(𝑛 +1)(2𝑛 +1))/6 " + " (𝑛(𝑛 + 1))/2) = 1/6 (𝑛2(𝑛 + 1)2/2 " + " (𝑛(𝑛 +1)(2𝑛 +1))/2 " + " (𝑛(𝑛 + 1))/2) = 1/6 ((𝑛(𝑛 + 1))/2) (n(n + 1) + (2n + 1) + 1) = (𝑛(𝑛 + 1))/12[n(n + 1) + (2n + 1) + 1] = (𝑛(𝑛 + 1))/12[n2 + n + 2n + 1 + 1] = (𝑛(𝑛 + 1))/12[n2 + 3n + 2] = (𝑛(𝑛 + 1))/12[n2 + 2n + n + 2] = (𝑛(𝑛 + 1))/12[n(n + 2) + (n + 2)] = (𝑛(𝑛 + 1))/12[(n + 1)(n + 2)] = (𝑛(𝑛 + 1)2(𝑛 + 2))/12 Thus ,the required sum is (𝑛(𝑛 + 1)2(𝑛 + 2))/12

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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.