Question

Find the sum of the series $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+------------------------------------12terms$

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Answer 1

Let $T_n$ be the $n^{th}$ term of the given series

To get the $T_n$ of the given series, separetely find the $n^{th}$ term of numerator and denominator and then divide them.

$T_n$ = $\frac{1^3+2^3+3^3+----------------n^3}{(1+3+5+--------------2n-1)}$ = $\frac{n^2(n+1{)}^2}{\frac{4}{n^2}}$

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

Sum of the n terms $S_n$ = ${\sum }_{i=1}^n$ $T_n$

= ${\sum }_{i=1}^n$ $\frac{1}{4}$ ($n^2$ + 1 +2n)

= $\frac{1}{4}$ $\left[{\sum }_{i=1}^n{n}^2+{\sum }_{i=1}^{n}1+{\sum }_{i=1}^{n}2n\right]$

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

= $\frac{n}{24}$ {2$n^2$ + 3n + 1 + 6n + 6 + 6}

= $\frac{n(2n^2+9n+18)}{24}$-----------------(1)

Substitution n = 12 is equation 1

$S_{12}$ = $\frac{12\times (2\times 144+9\times 12+18)}{24}$

= 207