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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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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} $

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1 Answer

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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]$
$\Rightarrow\:S_n=\large\frac{n(2n^2+9n+13)}{24}$
answered Apr 6, 2014 by rvidyagovindarajan_1
 

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