Answer

Verified

427.2k+ 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.

Recently Updated Pages

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Which one of the following places is not covered by class 10 social science CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

The states of India which do not have an International class 10 social science CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

How do you graph the function fx 4x class 9 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

Name the three parallel ranges of the Himalayas Describe class 9 social science CBSE