\\sum _{k = 1 } ^{n} ( 2 ^ { k - 1 } + 8 k ^ { 3 } - 6 k ^ { 2 } ) =

Correct Answer is : (2n - 1) + n (n + 1) (2n2 - 1)

TS ICET 2017 Question Paper with Solutions for online practice

Question : 129 of 200

Marks: +1, -0

\Σ ↙{k = 1 } ↖{n} ( 2 ^ { k - 1 } + 8 k ^ { 3 } - 6 k ^ { 2 } ) =

Solution:

∑↙{k=1}↖{n}2^{k-1}+8k^{3}-6k^{2}

=∑↙{k=1}↖{n}2^{k-1}+8∑↙{k=1}↖{n}k^{3}-6∑↙{k=1}↖{n}k^{2}

=(2^{n}-1)+8\{{n(n+1)}/{2}\}^{2}-6×{n(n+1)(2n+1)}/{6}

=(2^{n}-1)+2n^{2}(n+1)-n(n+1)(2n+1)

=(2^{n}-1)+n(n+1)\{2n(n+1)-(2n+1)\}

=(2^{n}-1)+n(n+1)(2n^{2}-1)

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