Find the sum of the series (2^3-1)+(4^3-3^3)+(6^3-15^3)......n terms
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(8-1) +(64-27)+(216-3375)
7+37-3159
44-3159
-3115
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we have to find the sum of series,
(2³ - 1³) + (4³ - 3³) + (6³ - 5³) + ...... + {(2n)³ - (2n-1)³}
so, sum of series,
= [(2n)³ - (2n - 1)³ ]
= [8n³ - 8n³ + 12n² - 6n + 1 ]
=
= 12{n(n + 1)/2}² - 6n(n + 1)/2 + n
= 3n²(n + 1)² -3n(n + 1) + n
= 3n(n + 1) [n(n + 1) - 1] + n
= 3n(n + 1) [n² + n - 1] + n
= 3n(n + 1)(n² + n - 1) + n
hence sum of the series is 3n(n + 1)(n² + n - 1) + n
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