Class 11 Mathematics - Chapter Sequence and Series NCERT Solutions | Find the sum to n terms of the

Welcome to the NCERT Solutions for Class 11th Mathematics - Chapter Sequence and Series. This page offers a step-by-step solution to the specific question from Excercise ".$ex_no." , Question 5: find the sum to nbsp n nbsp terms of the series 52....
Question 5

Find the sum to n terms of the series 52 + 62 + 72 + ... + 202

Answer

The given series is 52 + 62 + 72 + … + 202

nth term, an = ( n + 4)2 = n2 + 8n + 16

therefore space S subscript n space equals space sum from k equals 1 to n of space a subscript k space equals space sum from k equals 1 to n of space open parentheses k squared space plus space 8 k space plus space 16 close parentheses
space space space space space space space space space space space space equals space sum from k equals 1 to n of space k squared space plus space 8 sum from k equals 1 to n of space k space plus space sum from k equals 1 to n of space 16
space space space space space space space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction space plus fraction numerator 8 n open parentheses n plus 1 close parentheses over denominator 2 end fraction space plus 16 n

16th term is (16 + 4)2 = (20)2

therefore space S subscript n space equals space fraction numerator 16 open parentheses 16 plus 1 close parentheses open parentheses 2 cross times 16 plus 1 close parentheses over denominator 6 end fraction plus fraction numerator 8 cross times 16 cross times open parentheses 16 plus 1 close parentheses over denominator 2 end fraction plus 16 cross times 16
space space space space space space space space space space space equals space space fraction numerator open parentheses 16 close parentheses open parentheses 17 close parentheses open parentheses 33 close parentheses over denominator 6 end fraction plus fraction numerator open parentheses 8 close parentheses cross times 16 cross times open parentheses 16 plus 1 close parentheses over denominator 2 end fraction plus 16 cross times 16
space space space space space space space space space space space equals space space fraction numerator open parentheses 16 close parentheses open parentheses 17 close parentheses open parentheses 33 close parentheses over denominator 6 end fraction plus fraction numerator open parentheses 8 close parentheses open parentheses 16 close parentheses open parentheses 17 close parentheses over denominator 2 end fraction plus 256
space space space space space space space space space space equals 1496 space plus space 1088 space plus space 256
space space space space space space space space space space equals space 2840
therefore space 5 squared space plus space 6 to the power of 2 space end exponent plus space 7 squared space plus space...... space plus space 20 squared space equals space 2840

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