99. Find the sum of the following series up to n terms: 131+13+231+3+13+23+331+3+5+…

99. The given series is. 131+13+231+3+13+23+331+3+5 +………….. upto nth term, The nth term is , an=13+23+33+........+n31+3+5+........+n...terms { A.P. of n terms and a = 1, d = 5-3 = 2} an=[n(n+1)2]2n2[2*1+(n−1)2] an=n2(n+1)24n2[2+(n−1)2]=n2(n+1)24n[1+n−1] an=n2(n+1)24n2=(n+1)24 an=n2+2x+14=n24+n2+14 ∴Sn=∑an=14∑x2+12∑n+∑1 =14*9(n+1)(2n+1)6+12*n(x+1)2+14*n =n4[(n+1)(2n+1)6+(n+1)+1] =n4[2n2+n(n+1)(2n+1)+6(x+1)+66] =n4[2n2+n+2n+1+6n+6+66] =n4[2n2+9n+136] =n(2n2+9n+13)24

99. Find the sum of the following series up to n terms: $\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + \dots$

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Payal Gupta | Contributor-Level 10

9 months ago

99. The given series is.

$\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5}$ +………….. upto nth term,

The nth term is ,

$a_{n} = \frac{1^{3} + 2^{3} + 3^{3} + . . . . . . . . + n^{3}}{1 + 3 + 5 + . . . . . . . . + n . . . t e r m s}$ { A.P. of n terms and a = 1, d = 5-3 = 2}

$a_{n} = \frac{{[\frac{n (n + 1)}{2}]}^{2}}{\frac{n}{2} [2 * 1 + (n - 1) 2]}$

$a_{n} = \frac{\frac{n^{2} (n + 1)^{2}}{4}}{\frac{n}{2} [2 + (n - 1) 2]} = \frac{\frac{n^{2} (n + 1)^{2}}{4}}{n [1 + n - 1]}$

$a_{n} = \frac{\frac{n^{2} (n + 1)^{2}}{4}}{n^{2}} = \frac{{(n + 1)}^{2}}{4}$

$a_{n} = \frac{n^{2} + 2 x + 1}{4} = \frac{n^{2}}{4} + \frac{n}{2} + \frac{1}{4}$

$∴ S n = \sum_{}^{} a n = \frac{1}{4} \sum_{}^{} x^{2} + \frac{1}{2} \sum_{}^{} n + \sum_{}^{} 1$

$= \frac{1}{4} * \frac{9 (n + 1) (2 n + 1)}{6} + \frac{1}{2} * \frac{n (x + 1)}{2} + \frac{1}{4} * n$

$= \frac{n}{4} [\frac{(n + 1) (2 n + 1)}{6} + (n + 1) + 1]$

$= \frac{n}{4} [\frac{2 n^{2} + n (n + 1) (2 n + 1) + 6 (x + 1) + 6}{6}]$

$= \frac{n}{4} [\frac{2 n^{2} + n + 2 n + 1 + 6 n + 6 + 6}{6}]$

$= \frac{n}{4} [\frac{2 n^{2} + 9 n + 13}{6}]$

$= \frac{n (2 n^{2} + 9 n + 13)}{24}$

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