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# Find the sum of the following series $1+6+\dfrac{9\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}} \right)}{7}+\dfrac{12\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}} \right)}{9}+\dfrac{15\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{5}^{2}} \right)}{11}+......$ up to 15 terms?(a) 7820(b) 7830(c) 7520(d) 7510

Last updated date: 13th Jun 2024
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Hint:
We start solving the problem by making the simplifications in the given series to find the general term that represents every term of the series. Once we find the general terms, we make the simplifications in it by making use of the fact that sum of squares of first n natural numbers is defined as ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ . We then substitute this in sum and make use of the sum of squares and cubes of first n natural numbers to get the required answer.

According to the problem, we are asked to find the sum of the given series $1+6+\dfrac{9\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}} \right)}{7}+\dfrac{12\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}} \right)}{9}+\dfrac{15\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{5}^{2}} \right)}{11}+......$ up to 15 terms.
Let us assume $S=1+6+\dfrac{9\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}} \right)}{7}+\dfrac{12\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}} \right)}{9}+\dfrac{15\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{5}^{2}} \right)}{11}+......$ .
$\Rightarrow S=\dfrac{3\left( 1 \right)}{3}+\dfrac{6\left( 5 \right)}{5}+\dfrac{9\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}} \right)}{7}+\dfrac{12\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}} \right)}{9}+\dfrac{15\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{5}^{2}} \right)}{11}+......$ .
$\Rightarrow S=\dfrac{3\left( 1 \right)}{3}+\dfrac{6\left( 1+4 \right)}{5}+\dfrac{9\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}} \right)}{7}+\dfrac{12\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}} \right)}{9}+\dfrac{15\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{5}^{2}} \right)}{11}+......$ .
$\Rightarrow S=\dfrac{3\left( {{1}^{2}} \right)}{3}+\dfrac{6\left( {{1}^{2}}+{{2}^{2}} \right)}{5}+\dfrac{9\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}} \right)}{7}+\dfrac{12\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}} \right)}{9}+\dfrac{15\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{5}^{2}} \right)}{11}+......$ .
$\Rightarrow S=\dfrac{\left( 3\times 1 \right)\left( {{1}^{2}} \right)}{\left( 2+1 \right)}+\dfrac{\left( 3\times 2 \right)\left( {{1}^{2}}+{{2}^{2}} \right)}{\left( 4+1 \right)}+\dfrac{\left( 3\times 3 \right)\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}} \right)}{\left( 6+1 \right)}+\dfrac{\left( 3\times 4 \right)\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}} \right)}{\left( 8+1 \right)}+\dfrac{\left( 3\times 5 \right)\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{5}^{2}} \right)}{\left( 10+1 \right)}+......$.
$\Rightarrow S=\dfrac{\left( 3\times 1 \right)\left( {{1}^{2}} \right)}{\left( \left( 2\times 1 \right)+1 \right)}+\dfrac{\left( 3\times 2 \right)\left( {{1}^{2}}+{{2}^{2}} \right)}{\left( \left( 2\times 2 \right)+1 \right)}+\dfrac{\left( 3\times 3 \right)\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}} \right)}{\left( \left( 2\times 3 \right)+1 \right)}+\dfrac{\left( 3\times 4 \right)\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}} \right)}{\left( \left( 2\times 4 \right)+1 \right)}+\dfrac{\left( 3\times 5 \right)\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{5}^{2}} \right)}{\left( \left( 2\times 5 \right)+1 \right)}+......$ ---(1).
From equation (1), we can see that each term of the series is in the form of ${{T}_{r}}=\dfrac{\left( 3r \right)\left( {{1}^{2}}+{{2}^{2}}+....+{{r}^{2}} \right)}{\left( 2r+1 \right)}$ .
So, we can write equation (1) as $S=\sum\limits_{r=1}^{r=15}{{{T}_{r}}}$ ---(2).
Let us first simplify ${{T}_{r}}$ and then substitute that in equation (2).
So, we have ${{T}_{r}}=\dfrac{\left( 3r \right)\left( {{1}^{2}}+{{2}^{2}}+....+{{r}^{2}} \right)}{\left( 2r+1 \right)}$ .
We know that the sum of squares of first n natural numbers is defined as ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ .
$\Rightarrow {{T}_{r}}=\dfrac{\left( 3r \right)\left( \dfrac{r\left( r+1 \right)\left( 2r+1 \right)}{6} \right)}{\left( 2r+1 \right)}$ .
$\Rightarrow {{T}_{r}}=\dfrac{{{r}^{2}}\left( r+1 \right)}{2}$ .
$\Rightarrow {{T}_{r}}=\dfrac{{{r}^{3}}+{{r}^{2}}}{2}$ ---(3).
Let us substitute equation (3) in equation (2).
$\Rightarrow S=\sum\limits_{r=1}^{r=15}{\left( \dfrac{{{r}^{3}}+{{r}^{2}}}{2} \right)}$ .
$\Rightarrow S=\dfrac{1}{2}\left( \sum\limits_{r=1}^{r=15}{{{r}^{3}}}+\sum\limits_{r=1}^{r=15}{{{r}^{2}}} \right)$ .
We know that sum of squares of first n natural numbers is defined as ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ and sum of cubes of first n natural numbers is defined as ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+....+{{n}^{3}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}$ .
$\Rightarrow S=\dfrac{1}{2}{{\left( \dfrac{{{r}^{2}}{{\left( r+1 \right)}^{2}}}{4}+\dfrac{r\left( r+1 \right)\left( 2r+1 \right)}{6} \right)}_{r=15}}$ .
$\Rightarrow S=\dfrac{1}{2}\left( \dfrac{{{15}^{2}}{{\left( 16 \right)}^{2}}}{4}+\dfrac{15\left( 16 \right)\left( 31 \right)}{6} \right)$ .
$\Rightarrow S=\dfrac{1}{2}\left( 14400+1240 \right)$ .
$\Rightarrow S=\dfrac{1}{2}\left( 15640 \right)$ .
$\Rightarrow S=7820$ .
So, we have found the sum of the given series up to 15 terms as 7820.
$\therefore,$ The correct option for the given problem is (a).

Note:
We can see that the given problem contains a huge amount of calculation, so we need to perform each step carefully in order to avoid confusion and calculation mistakes. Whenever we get this type of problem, we first find the general terms to represent every term of the series as we can see that this general leads us to the required solution. Similarly, we can expect problems to find the sum to 100 terms of the given series.