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Ex 9.4, 5 - Find sum of series 52 + 62 + 72 + ... + 202 - Finding sum from nth number

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
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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) = (𝑛(𝑛+1)(2𝑛+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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