Permutations And Combinations

Question
CBSEENMA11013157

Find the sum of n terms of the series whose nth term is given by 

straight n squared plus 2 to the power of straight n

Solution

Here, straight T subscript straight n space equals space straight n squared plus 2 to the power of straight n
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space space space equals left parenthesis 1 squared plus 2 squared plus 3 squared plus.... plus straight n squared right parenthesis plus left parenthesis 2 to the power of 1 plus 2 squared plus.....2 to the power of straight n right parenthesis space equals space sum straight n squared plus fraction numerator 2 left parenthesis 2 to the power of straight n minus 1 right parenthesis over denominator 2 minus 1 end fraction
                            {The series is 2nd bracket is G.P. with common ratio 2>1}
                     equals space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis left parenthesis 2 straight n plus 1 right parenthesis over denominator 6 end fraction space plus 2 to the power of straight n plus 1 end exponent minus 2

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Some More Questions From Permutations and Combinations Chapter

Determine K, so that K + 2, 4K – 6 and 3K – 2 are three consecutive terms of an A.P.