Browse Questions

# Find the sum of $n$ terms of the series $\large\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+.........$

$\begin{array}{1 1} \large\frac{n(2n^2+9n+13)}{24} \\\large\frac{n(2n^2+9n+1)}{24} \\ \large\frac{n(2n^2+9n+13)}{12} \\\large\frac{n(2n^2+10n+13)}{24}\end{array}$

Toolbox:
• $\sum n^3=\large\frac{n^2(n+1)^2}{4}$
• Sum of $n$ terms of an A.P. $=\large\frac{n}{2}$$\big[2a+(n-1)d\big] • Sum of n terms of any series =S_n=\sum t_n • \sum n^2=\large\frac{n(n+1)(2n+1)}{6} • \sum n=\large\frac{n(n+1)}{2} • \sum 1=n Step 1 Given series is \large\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+......... We have to find the n^{th} term 't_n'of the series . The numerator of the n^{th} term is 1^3+2^3+3^3+.......=\sum n^3 But we know that \sum n^3=\large\frac{n^2(n+1)^2}{4} \therefore The numerator of t_n is \large\frac{n^2(n+1)^2}{4}.....(i) Step 2 The denominator of t_n is 1+3+5+........ This series is an A.P. with 1^{st} term= a=1 and common difference d=2 We know that sum of n terms of an A.P. =\large\frac{n}{2}$$\big[2a+(n-1)d\big]$
$\therefore$ $1+3+5+......=\large\frac{n}{2}$$[2\times 1+(n-1).2\big] \qquad\:=\large\frac{n}{2}.2n=n^2 i.e., The denominator of t_n=n^2.......(ii) Step3 Now t_n=\large\frac{n^2(n+1)^2}{4n^2} =\large\frac{(n+1)^2}{4}$$=\large\frac{1}{4}$$(n^2+2n+1) We know that sum of n terms of any series =S_n=\sum t_n \therefore Sum of n terms of the given series is =S_n= \sum\large\frac{1}{4}$$(n^2+2n+1)$
$=\large\frac{1}{4}$$\bigg[\sum n^2+2\sum n+\sum 1\bigg] We know that \sum n^2=\large\frac{n(n+1)(2n+1)}{6},\sum n=\large\frac{n(n+1)}{2} and \sum 1=n \Rightarrow\:S_n=\large\frac{1}{4}\bigg[\large\frac{n(n+1)(2n+1)}{6}$$+2.\large\frac{n(n+1)}{2}$$+n\bigg] \qquad\:=\large\frac{n}{4}$$\bigg[\large\frac{(n+1)(2n+1)+6n+6+6}{6}\bigg]$
$\Rightarrow\:S_n=\large\frac{n(2n^2+9n+13)}{24}$