Math, asked by Mugundhan, 11 months ago

Find the sum of the series (2^3-1)+(4^3-3^3)+(6^3-15^3)......n terms

Answers

Answered by Sandeepyaduvanshi
2

(8-1) +(64-27)+(216-3375)

7+37-3159

44-3159

-3115

Answered by abhi178
5

we have to find the sum of series,

(2³ - 1³) + (4³ - 3³) + (6³ - 5³) + ...... + {(2n)³ - (2n-1)³}

so, sum of series, S_n=\Sigma T_n

= \Sigma[(2n)³ - (2n - 1)³ ]

= \Sigma [8n³ - 8n³ + 12n² - 6n + 1 ]

= \Sigma 12n^2-\Sigma 6n+\Sigma 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

also read similar questions : 5/6+3+3/4<br /> \frac{5}{6} + 3 + \frac{3}{4} <br />find the sum

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the sum of n terms of the series 2,5,8,11 is..60100 ,then n is?

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