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# Find the sum of the series $\dfrac{3\times {{1}^{3}}}{{{1}^{2}}}+\dfrac{5\times \left( {{1}^{3}}+{{2}^{3}} \right)}{{{1}^{2}}+{{2}^{2}}}+\dfrac{7\times \left( {{1}^{3}}+{{2}^{3}}+{{3}^{3}} \right)}{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}+.......$ up to 10 terms.(a) 660(b) 620(c) 680(d) 600

Last updated date: 18th Jun 2024
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Hint: We start solving the problem by finding the general equation to represent each term of the series. After finding the general equation, we take summation of it up to n terms. Once we find the summation up to n terms, we substitute 10 in place in n and make required calculations to get the desired result.

According to the problem, we need to find the sum of the series $\dfrac{3\times {{1}^{3}}}{{{1}^{2}}}+\dfrac{5\times \left( {{1}^{3}}+{{2}^{3}} \right)}{{{1}^{2}}+{{2}^{2}}}+\dfrac{7\times \left( {{1}^{3}}+{{2}^{3}}+{{3}^{3}} \right)}{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}+.......$ up to 10 terms.
Let us find the general term of the series to solve for the sum of the series.
$\Rightarrow \dfrac{\left( 2+1 \right)\times {{1}^{3}}}{{{1}^{2}}}+\dfrac{\left( 4+1 \right)\times \left( {{1}^{3}}+{{2}^{3}} \right)}{{{1}^{2}}+{{2}^{2}}}+\dfrac{\left( 6+1 \right)\times \left( {{1}^{3}}+{{2}^{3}}+{{3}^{3}} \right)}{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}+.......$ .
$\Rightarrow \dfrac{\left( 2\left( 1 \right)+1 \right)\times {{1}^{3}}}{{{1}^{2}}}+\dfrac{\left( 2\left( 2 \right)+1 \right)\times \left( {{1}^{3}}+{{2}^{3}} \right)}{{{1}^{2}}+{{2}^{2}}}+\dfrac{\left( 2\left( 3 \right)+1 \right)\times \left( {{1}^{3}}+{{2}^{3}}+{{3}^{3}} \right)}{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}+.......$ ---(1).
We can see that each term is of the form $\dfrac{\left( 2r+1 \right)\times \sum\limits_{1}^{r}{{{r}^{3}}}}{\sum\limits_{1}^{r}{{{r}^{2}}}}$ for $r=1,2,3,......n$.
We know that sum of the squares of the first n natural numbers is $\sum\limits_{r=1}^{n}{{{r}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}}$ and sum of the cubes of the first n natural numbers is $\sum\limits_{r=1}^{n}{{{r}^{3}}}=\dfrac{{{n}^{2}}\times {{\left( n+1 \right)}^{2}}}{4}$.
So, we get general term as $\dfrac{\left( 2r+1 \right)\times \sum\limits_{1}^{r}{{{r}^{3}}}}{\sum\limits_{1}^{r}{{{r}^{2}}}}=\dfrac{\left( 2r+1 \right)\times \left( \dfrac{{{r}^{2}}\times {{\left( r+1 \right)}^{2}}}{4} \right)}{\left( \dfrac{r\times \left( r+1 \right)\times \left( 2r+1 \right)}{6} \right)}$.
$\Rightarrow \dfrac{\left( 2r+1 \right)\times \sum\limits_{1}^{r}{{{r}^{3}}}}{\sum\limits_{1}^{r}{{{r}^{2}}}}=\dfrac{\left( \dfrac{r\times \left( r+1 \right)}{2} \right)}{\left( \dfrac{1}{3} \right)}$.
$\Rightarrow \dfrac{\left( 2r+1 \right)\times \sum\limits_{1}^{r}{{{r}^{3}}}}{\sum\limits_{1}^{r}{{{r}^{2}}}}=\dfrac{3}{2}\times \left( r\times \left( r+1 \right) \right)$.
$\Rightarrow \dfrac{\left( 2r+1 \right)\times \sum\limits_{1}^{r}{{{r}^{3}}}}{\sum\limits_{1}^{r}{{{r}^{2}}}}=\dfrac{3}{2}\times \left( {{r}^{2}}+r \right)$.
We can represent sum of the series in equation (1) as $\sum\limits_{r=1}^{n}{\dfrac{3}{2}\times \left( {{r}^{2}}+r \right)}$.
$\Rightarrow \sum\limits_{r=1}^{n}{\dfrac{3}{2}\times \left( {{r}^{2}}+r \right)}=\dfrac{3}{2}\times \left( \sum\limits_{r=1}^{n}{{{r}^{2}}}+\sum\limits_{r=1}^{n}{r} \right)$ ---(2).
We know that sum of the squares of the first n natural numbers is defined as $\sum\limits_{r=1}^{n}{{{r}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}}$ and sum of the first n natural numbers is defined as $\sum\limits_{r=1}^{n}{r}=\dfrac{n\left( n+1 \right)}{2}$. We use these results in equation (2).
$\Rightarrow \sum\limits_{r=1}^{n}{\dfrac{3}{2}\times \left( {{r}^{2}}+r \right)}=\dfrac{3}{2}\times \left( \left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right)+\left( \dfrac{n\left( n+1 \right)}{2} \right) \right)$.
$\Rightarrow \sum\limits_{r=1}^{n}{\dfrac{3}{2}\times \left( {{r}^{2}}+r \right)}=\left( \dfrac{3\times n\times \left( n+1 \right)}{2\times 2} \right)\times \left( \dfrac{\left( 2n+1 \right)}{3}+1 \right)$.
$\Rightarrow \sum\limits_{r=1}^{n}{\dfrac{3}{2}\times \left( {{r}^{2}}+r \right)}=\left( \dfrac{3\times n\times \left( n+1 \right)}{4} \right)\times \left( \dfrac{2n+1+3}{3} \right)$.
$\Rightarrow \sum\limits_{r=1}^{n}{\dfrac{3}{2}\times \left( {{r}^{2}}+r \right)}=\left( \dfrac{3\times n\times \left( n+1 \right)}{4} \right)\times \left( \dfrac{2n+4}{3} \right)$.
$\Rightarrow \sum\limits_{r=1}^{n}{\dfrac{3}{2}\times \left( {{r}^{2}}+r \right)}=\left( \dfrac{3\times n\times \left( n+1 \right)}{4} \right)\times \left( \dfrac{2\left( n+2 \right)}{3} \right)$.
$\Rightarrow \sum\limits_{r=1}^{n}{\dfrac{3}{2}\times \left( {{r}^{2}}+r \right)}=\left( \dfrac{n\times \left( n+1 \right)\times \left( n+2 \right)}{2} \right)$ ---(3).
Now we substitute 10 in place of n in equation (3).
So, we have $\sum\limits_{r=1}^{10}{\dfrac{3}{2}\times \left( {{r}^{2}}+r \right)}=\left( \dfrac{10\times \left( 10+1 \right)\times \left( 10+2 \right)}{2} \right)$.
$\Rightarrow \sum\limits_{r=1}^{10}{\dfrac{3}{2}\times \left( {{r}^{2}}+r \right)}=\left( \dfrac{10\times \left( 11 \right)\times \left( 12 \right)}{2} \right)$.
$\Rightarrow \sum\limits_{r=1}^{10}{\dfrac{3}{2}\times \left( {{r}^{2}}+r \right)}=10\times 11\times 6$.
$\Rightarrow \sum\limits_{r=1}^{10}{\dfrac{3}{2}\times \left( {{r}^{2}}+r \right)}=660$.
We have found the sum of the series $\dfrac{3\times {{1}^{3}}}{{{1}^{2}}}+\dfrac{5\times \left( {{1}^{3}}+{{2}^{3}} \right)}{{{1}^{2}}+{{2}^{2}}}+\dfrac{7\times \left( {{1}^{3}}+{{2}^{3}}+{{3}^{3}} \right)}{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}+.......$ up to 10 terms as 660.
∴ The sum of the series $\dfrac{3\times {{1}^{3}}}{{{1}^{2}}}+\dfrac{5\times \left( {{1}^{3}}+{{2}^{3}} \right)}{{{1}^{2}}+{{2}^{2}}}+\dfrac{7\times \left( {{1}^{3}}+{{2}^{3}}+{{3}^{3}} \right)}{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}+.......$ up to 10 terms as 660.
The correct option for the given problem is (a).

Note: We should the take the general equation to represent each term of series as $\dfrac{\left( 2r+1 \right).{{r}^{3}}}{{{r}^{2}}}$, because it will accommodate cube and square of only one number. Whenever we get this type of problem, we try to find the general equation of the terms which makes our sum easier. We can also take 10 in place of n while finding the general summation for n terms.