Question

Find the sum of the series whose ${{n}^{th}}$ term is ${{n}^{2}}\left( {{n}^{2}}-1 \right)$.

Answer 1

Hint: Use the summation of special series \[\sum{{{n}^{2}}}\] and $\sum{{{n}^{4}}}$ to get the sum of $n$ terms of the given series. 

Complete Step-by-step answer:

Let us suppose the given ${{n}^{th}}$term i.e., ${{n}^{2}}\left( {{n}^{2}}-1 \right)$ be ${{T}_{n}}$. Hence,

${{T}_{n}}={{n}^{2}}\left( {{n}^{2}}-1 \right)$…………………………………………..(i)

Now, for getting sum of this series we can apply summation to given such as

$\sum\limits_{n=1}^{n}{{{T}_{n}}}=\sum\limits_{n=1}^{n}{{{n}^{2}}\left( {{n}^{2}}-1 \right)}$………………………………………….(ii)

Series can be given as

${{1}^{2}}\left( {{1}^{2}}-1 \right)+{{2}^{2}}\left( {{2}^{2}}-1 \right)+{{3}^{2}}\left( {{3}^{2}}-1 \right)+{{4}^{2}}\left( {{4}^{2}}-1 \right)+..............{{n}^{2}}\left( {{n}^{2}}-1 \right)$

Let us represent this series with ${{S}_{n}}$. Hence, we get

${{S}_{n}}={{1}^{2}}\left( {{1}^{2}}-1 \right)+{{2}^{2}}\left( {{2}^{2}}-1 \right)+{{3}^{2}}\left( {{3}^{2}}-1 \right)+{{4}^{2}}\left( {{4}^{2}}-1 \right)+..............{{n}^{2}}\left( {{n}^{2}}-1 \right)$

$\Rightarrow {{S}_{n}}=\left( {{1}^{4}}-{{1}^{2}} \right)+\left( {{2}^{4}}-{{2}^{2}} \right)+\left( {{3}^{4}}-{{3}^{2}} \right)+\left( {{4}^{4}}-{{4}^{2}} \right)+.............\left( {{n}^{4}}-{{n}^{2}} \right)$

Now, we can simplify the above series by taking ${{1}^{4}},{{2}^{4}},{{3}^{4}}.........{{n}^{4}}$ in one bracket and ${{1}^{2}},{{2}^{2}},{{3}^{2}}.........{{n}^{2}}$in another; Hence, we get

${{S}_{n}}=\left( {{1}^{4}}+{{2}^{4}}+{{3}^{4}}+.........{{n}^{4}} \right)-\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.........{{n}^{2}} \right)$……………………………….(iii)

Now, we know the identities of summation of special series as

$\sum\limits_{n=1}^{n}{{{n}^{2}}}={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.........{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$

$\sum\limits_{n=1}^{n}{n}=1+2+3+........n=\dfrac{n\left( n+1 \right)}{2}$

$\sum\limits_{n=1}^{n}{{{n}^{3}}}={{1}^{3}}+{{2}^{3}}+{{3}^{3}}+........{{n}^{3}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}$

$\sum\limits_{n=1}^{n}{{{n}^{4}}}={{1}^{4}}+{{2}^{4}}+{{3}^{4}}+........{{n}^{4}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)\left( 3{{n}^{2}}+3n-1 \right)}{30}$

So, we can put values of $\sum{{{n}^{4}}}$and $\sum{{{n}^{2}}}$from above equations to equations (iii); Hence, we get

${{S}_{n}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)\left( 3{{n}^{2}}+3n-1 \right)}{30}-\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$

Here, we can take $\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ as common for further simplify the given equation in ${{S}_{n}}$. Hence, we get

${{S}_{n}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\left[ \dfrac{3{{n}^{2}}+3n-1}{5}-\dfrac{1}{1} \right]$${{S}_{n}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\left[ \dfrac{3{{n}^{2}}+3n-1}{5}-\dfrac{1}{1} \right]$

Taking L.C.M. in bracket, we get

${{S}_{n}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\left[ \dfrac{3{{n}^{2}}+3n-1-5}{5} \right]$

$\Rightarrow {{S}_{n}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)\left( 3{{n}^{2}}+3n-6 \right)}{30}$

Now, taking ‘3’ as common from $3{{n}^{2}}+3n-6$, we get

${{S}_{n}}=\dfrac{3n\left( n+1 \right)\left( 2n+1 \right)\left( {{n}^{2}}+n-2 \right)}{30}$

$\Rightarrow {{S}_{n}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)\left( {{n}^{2}}+n-2 \right)}{10}$………………………………………….(iv)

Now, we can factorize ${{n}^{2}}+n-2$ by splitting middle term ‘$n$’ as ‘$2n-n$’ to simplify ‘${{S}_{n}}$’ further:-

$ {{n}^{2}}+n-2={{n}^{2}}+2n-n-2 $

 $ \Rightarrow {{n}^{2}}+n-2=n\left( n+2 \right)-1\left( n+2 \right) $

 $ \Rightarrow {{n}^{2}}+n-2=\left( n-1 \right)\left( n+2 \right) $

Hence, equation (iv) can be further simplified as

${{S}_{n}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)\left( n-1 \right)\left( n+2 \right)}{10}$

Now, we can replace $\left( n-1 \right)\left( n+1 \right)$ by ${{n}^{2}}-1$ using algebraic identity $\left( a-b \right)\left( a+b \right)=\left( {{a}^{2}}-{{b}^{2}} \right)$ . Hence, we get

${{S}_{n}}=\dfrac{n\left( n+2 \right)\left( 2n+1 \right)\left( {{n}^{2}}-1 \right)}{10}$

Note: As we generally do not use the special series ${{1}^{4}}+{{2}^{4}}+{{3}^{4}}+.......{{n}^{4}}$ in the problems of sequence and series chapter.
One can prove the summation of $\sum{{{n}^{4}}}$ as follows: -
We have
${{\left( n+1 \right)}^{5}}-{{n}^{5}}=5{{n}^{4}}-10{{n}^{3}}+10{{n}^{2}}-5n+1$
Now, put $n=1,2,3,....n$ and add all the equations to get summation as
${{2}^{5}}-{{1}^{5}}=5\cdot {{1}^{4}}-10\cdot {{1}^{3}}+10\cdot {{1}^{2}}-5\cdot 1+1$
${{3}^{5}}-{{2}^{5}}=5\cdot {{2}^{4}}-10\cdot {{2}^{3}}+10\cdot {{2}^{2}}-5\cdot 2+1$
${{4}^{5}}-{{3}^{5}}=5\cdot {{3}^{4}}-10\cdot {{3}^{3}}+10\cdot {{2}^{3}}-5\cdot 3+1$
Till ‘n’ terms
Now, add all these equations and use $\sum{{{n}^{2}}}$ and $\sum{{{n}^{3}}}$ to get $\sum{{{n}^{4}}}$.
Using a direct sum of special series always makes the solution flexible and easier.