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
Answer
Verified550.5k+ views
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.
Complete step by step 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.
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.
Complete step by step 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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success
Master Class 12 Maths: Engaging Questions & Answers for Success
Master Class 12 Biology: Engaging Questions & Answers for Success
Master Class 12 Physics: Engaging Questions & Answers for Success
Master Class 8 Maths: Engaging Questions & Answers for Success
Class 8 Question and Answer - Your Ultimate Solutions Guide
Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE
What is the median of the first 10 natural numbers class 10 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Which of the following does not have a fundamental class 10 physics CBSE
State and prove converse of BPT Basic Proportionality class 10 maths CBSE