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SEQUENCES AND SERIES — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSSEQUENCES AND SERIES3 Marks
Question
Find the sum to n terms in each of the series in whose $n^{th}$ terms is given by $(2n - 1)^2$
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Answer

Given: $a_n = (2n - 1)^2 = 4n^2 - 4n + 1 \therefore {S_n} = \sum\limits_{k = 1}^n {{a_k}} \sum\limits_{k = 1}^n {4{k^2} - 4k + 1} $
$= (4.1^2 - 4.1 + 1) + (4.2^2 - 4.2 + 1) + ......... (4.n^2 - 4.n + 1)$
$= 4(1^2 + 2^2 + ........ + n^2) - 4(1 + 2 + 3 + ...... + n) + (1 + 1 + ....... n terms)$
$= {{4n(n + 1)(2n + 1)} \over 6} - {{4n(n + 1)} \over 2} + n$
$= n\left( {{{2(n + 1)(2n + 1)} \over 3} - 2(n + 1) + 1} \right)$
$= n \left( \frac { 4 n ^ { 2 } + 6 n + 2 - 6 n - 6 + 3 } { 3 } \right)$
$= n \left( \frac { 4 n ^ { 2 } - 1 } { 3 } \right)$
$= \frac { n ( 2 n + 1 ) ( 2 n - 1 ) } { 3 }$

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