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 Exercise 4, Question 9: find the sum to nbsp n nbsp terms of the series wh....
Question 9

Find the sum to n terms of the series whose nth terms is given by n2 + 2n

Answer

an = n2 + 2n

therefore space S subscript n space equals space sum from k equals 1 to n of space k squared space plus space 2 to the power of k space equals space sum from k equals 1 to n of space k squared space space plus space sum from k equals 1 to n of space 2 to the power of k space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

C o n s i d e r space sum from k equals 1 to n of space 2 to the power of k space equals space 2 to the power of 1 space plus space 2 squared space plus 2 cubed space plus space...

The above series 2, 22, 23, … is a G.P. with both the first term and common ratio equal to 2.

therefore space sum from k equals 1 to n of space 2 to the power of k space equals space fraction numerator open parentheses 2 close parentheses open square brackets open parentheses 2 close parentheses to the power of n space minus 1 close square brackets over denominator 2 minus 1 end fraction equals 2 open parentheses 2 to the power of n space minus 1 close parentheses space space space space space space space space space space space... left parenthesis 2 right parenthesis

Therefore, from (1) and (2), we obtain

S subscript n space equals space sum from k equals 1 to n of space k squared space plus space 2 open parentheses 2 to the power of n space minus 1 close parentheses 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 plus 2 open parentheses 2 to the power of n space end exponent minus 1 close parentheses

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