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Hint: Find the nth term of the given series and apply $\sum{{}}$(summation) to the ${{T}_{n}}$ to get the sum of ‘n’ terms of the given series. Sum of special series can be given as $\sum{{{n}^{2}}=}\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ and $\sum{{{n}^{3}}=}\dfrac{n{{\left( n+1 \right)}^{2}}}{2}$and$\sum{n=}\dfrac{n\left( n+1 \right)}{2}$.

Complete step by step answer:

The given series up to n terms is

$\dfrac{{{1}^{3}}}{1}+\dfrac{{{1}^{3}}+{{2}^{3}}}{1+3}+\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}{1+3+5}+............$

Let the summation of the series up to ‘n’ terms be S, we get

$S=\dfrac{{{1}^{3}}}{1}+\dfrac{{{1}^{3}}+{{2}^{3}}}{1+3}+\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}{1+3+5}+$ ………………… (i)

As we know that we can find the sum of ‘n’ terms of the series by applying summation $\left( \sum{{}} \right)$ to the nth term of that series (general term).

So, first we need to find the value of the nth term of the given series.

Now, we can observe the series S in equation (i) and observe the numerator and denominator individually.

We get that numerator of first term has one term and numerator of second term has two term $\left( {{1}^{3}},{{2}^{3}} \right)$ and third term has three terms $\left( {{1}^{3}},{{2}^{3}},{{3}^{3}} \right)$and so on. So, nth term of the series should have ‘n’ terms in the numerator as $\left( {{1}^{3}},{{2}^{3}},{{3}^{3}}...........{{n}^{3}} \right)$.

Similarly, observe the denominator terms of the series S. So, we get that there is one term in the denominator in first term of the series, two terms (1, 3) in the second term of the series, and three terms (1, 3, 5) in the third term of the series and so on. So, the nth term of the series should have n terms in the denominator. That is it can be given as (1, 3, 5,……….).

Now (1, 3, 5,………..) sequence is an A.P. and nth term of an A.P is given as

${{T}_{n}}=a+\left( n-1 \right)d$

Here ‘a’ is the first term of A.P and d is the common difference of the A.P.

So, nth term of the sequence (1, 3, 5, ……….) can be given as

${{T}_{n}}=1+\left( n-1 \right)(2)=2n-1$

Hence, the nth term of the series ‘S’ has a denominator as (1, 3, 5,……..2n-1).

So, we can write the nth term of the series ‘S’ as

${{T}_{n}}=\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.......{{n}^{3}}}{1+3+5+..........\left( 2n-1 \right)}$ ……….. (ii)

${{T}_{n}}$is devoting the nth term of series S.

So, we can solve the numerator of equation (ii) by using the sum of special series of cube of natural numbers, given as

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

We can find value of denominator i.e., (1+3+5+……..2n-1) by applying the formula of sum of n terms of an A.P., given as

${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ ………………. (iv)

Here ‘n’ =number of terms, ‘a’= first term and ‘d’=common difference of A.P.

Therefore, 1+3+5+…………(2n-1) has a = 1, d = 2 and n = n

So, we get from equation (iv) as

$\left( 1+3+5+........2n-1 \right)=\dfrac{n}{2}\left[ 2\times 1+\left( n-1 \right)2 \right]=\dfrac{n}{2}\left[ 2+2n-2 \right]$

$\left( 1+3+5+........2n-1 \right)={{n}^{2}}$ ………… (v)

Now, put the values of equation (iii) and (v) in equation (ii) to simplify the nth term of series S, we get

$\begin{align}

& {{T}_{n}}=\dfrac{{{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}}{{{n}^{2}}} \\

& {{T}_{n}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4{{n}^{2}}} \\

\end{align}$

${{T}_{n}}=\dfrac{{{\left( n+1 \right)}^{2}}}{4}$………. (vi)

Now, we can apply summation sign $'\sum{'}$ to the expression of ${{T}_{n}}$ to get the summation of series ‘S’ as ‘${{T}_{n}}$’ is representing the nth term of the series ‘S’, we get

$\begin{align}

& S=\sum\limits_{n=1}^{n}{{{T}_{n}}} \\

& S=\sum\limits_{n=1}^{n}{\dfrac{\left( (n+1)^2 \right)}{4}} \\

& S=\dfrac{1}{4}\sum\limits_{n-1}^{n}{\left( {{n}^{2}}+1+2n \right)} \\

\end{align}$

$S=\dfrac{1}{4}\left[ \sum\limits_{n=1}^{n}{{{n}^{2}}+}\sum\limits_{n=1}^{n}{1}+2\sum\limits_{n=1}^{n}{n} \right]$…………… (vii)

Now, we know the sum of special series can be given by following formula:

$\begin{align}

& \sum\limits_{n=1}^{n}{n}={{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} \\

\end{align}$

\[\sum\limits_{n=1}^{n}{1}=1+1+1+......1=n\]

Hence, we can put the sum of series from the above formula in the equation (vii) to get the answer.

$S=\dfrac{1}{4}\left[ \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+n+2\dfrac{n\left( n+1 \right)}{2} \right]$

Taking out ‘n’, we get

\[\begin{align}

& S=\dfrac{n}{4}\left[ \dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}+1+\left( n+1 \right) \right] \\

& S=\dfrac{n}{4}\left[ \dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}+n+2 \right] \\

\end{align}\]

Taking the LCM, we get

$\begin{align}

& S=\dfrac{n}{4}\left[ \dfrac{\left( n+1 \right)\left( 2n+1 \right)+6n+12}{6} \right] \\

& S=\dfrac{n}{24}\left[ 2{{n}^{2}}+2n+n+1+6n+12 \right] \\

& S=\dfrac{n\left( 2{{n}^{2}}+9n+13 \right)}{24} \\

\end{align}$

Hence, the sum of series given in the problem is $\dfrac{n\left( 2{{n}^{2}}+9n+13 \right)}{24}$.

Note: We can find the sum by taking ${{T}_{n}}=\dfrac{{{\left( n+1 \right)}^{2}}}{4}$itself without expanding it in following ways.

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

Now put (n+1) in place of ‘n’ in the formula of sum of series $\sum{{{n}^{2}}}$ i.e., $\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ .

Hence, we will get

$\dfrac{1}{4}\left( \dfrac{\left( n+1 \right)\left( n+2 \right)\left( 2n+3 \right)}{6}-1 \right)$

We will get the same answer as found in the solution.

Don’t confuse it with the sign of $\sum{{}}$(summation). It represents a series in its nth term (general term) in very short form. So, it’s nothing but a representation of any series in a short way. And be clear with the summation of the special series $\sum{n},\sum{{{n}^{2}},}\sum{{{n}^{3}}}$ . Without the summation of these series, we will not be able to solve these kinds of series.

Getting ${{T}_{n}}$ and hence applying $\sum{{}}$ (summation) to ${{T}_{n}}$is the key point of the question.

Complete step by step answer:

The given series up to n terms is

$\dfrac{{{1}^{3}}}{1}+\dfrac{{{1}^{3}}+{{2}^{3}}}{1+3}+\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}{1+3+5}+............$

Let the summation of the series up to ‘n’ terms be S, we get

$S=\dfrac{{{1}^{3}}}{1}+\dfrac{{{1}^{3}}+{{2}^{3}}}{1+3}+\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}{1+3+5}+$ ………………… (i)

As we know that we can find the sum of ‘n’ terms of the series by applying summation $\left( \sum{{}} \right)$ to the nth term of that series (general term).

So, first we need to find the value of the nth term of the given series.

Now, we can observe the series S in equation (i) and observe the numerator and denominator individually.

We get that numerator of first term has one term and numerator of second term has two term $\left( {{1}^{3}},{{2}^{3}} \right)$ and third term has three terms $\left( {{1}^{3}},{{2}^{3}},{{3}^{3}} \right)$and so on. So, nth term of the series should have ‘n’ terms in the numerator as $\left( {{1}^{3}},{{2}^{3}},{{3}^{3}}...........{{n}^{3}} \right)$.

Similarly, observe the denominator terms of the series S. So, we get that there is one term in the denominator in first term of the series, two terms (1, 3) in the second term of the series, and three terms (1, 3, 5) in the third term of the series and so on. So, the nth term of the series should have n terms in the denominator. That is it can be given as (1, 3, 5,……….).

Now (1, 3, 5,………..) sequence is an A.P. and nth term of an A.P is given as

${{T}_{n}}=a+\left( n-1 \right)d$

Here ‘a’ is the first term of A.P and d is the common difference of the A.P.

So, nth term of the sequence (1, 3, 5, ……….) can be given as

${{T}_{n}}=1+\left( n-1 \right)(2)=2n-1$

Hence, the nth term of the series ‘S’ has a denominator as (1, 3, 5,……..2n-1).

So, we can write the nth term of the series ‘S’ as

${{T}_{n}}=\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.......{{n}^{3}}}{1+3+5+..........\left( 2n-1 \right)}$ ……….. (ii)

${{T}_{n}}$is devoting the nth term of series S.

So, we can solve the numerator of equation (ii) by using the sum of special series of cube of natural numbers, given as

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

We can find value of denominator i.e., (1+3+5+……..2n-1) by applying the formula of sum of n terms of an A.P., given as

${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ ………………. (iv)

Here ‘n’ =number of terms, ‘a’= first term and ‘d’=common difference of A.P.

Therefore, 1+3+5+…………(2n-1) has a = 1, d = 2 and n = n

So, we get from equation (iv) as

$\left( 1+3+5+........2n-1 \right)=\dfrac{n}{2}\left[ 2\times 1+\left( n-1 \right)2 \right]=\dfrac{n}{2}\left[ 2+2n-2 \right]$

$\left( 1+3+5+........2n-1 \right)={{n}^{2}}$ ………… (v)

Now, put the values of equation (iii) and (v) in equation (ii) to simplify the nth term of series S, we get

$\begin{align}

& {{T}_{n}}=\dfrac{{{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}}{{{n}^{2}}} \\

& {{T}_{n}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4{{n}^{2}}} \\

\end{align}$

${{T}_{n}}=\dfrac{{{\left( n+1 \right)}^{2}}}{4}$………. (vi)

Now, we can apply summation sign $'\sum{'}$ to the expression of ${{T}_{n}}$ to get the summation of series ‘S’ as ‘${{T}_{n}}$’ is representing the nth term of the series ‘S’, we get

$\begin{align}

& S=\sum\limits_{n=1}^{n}{{{T}_{n}}} \\

& S=\sum\limits_{n=1}^{n}{\dfrac{\left( (n+1)^2 \right)}{4}} \\

& S=\dfrac{1}{4}\sum\limits_{n-1}^{n}{\left( {{n}^{2}}+1+2n \right)} \\

\end{align}$

$S=\dfrac{1}{4}\left[ \sum\limits_{n=1}^{n}{{{n}^{2}}+}\sum\limits_{n=1}^{n}{1}+2\sum\limits_{n=1}^{n}{n} \right]$…………… (vii)

Now, we know the sum of special series can be given by following formula:

$\begin{align}

& \sum\limits_{n=1}^{n}{n}={{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} \\

\end{align}$

\[\sum\limits_{n=1}^{n}{1}=1+1+1+......1=n\]

Hence, we can put the sum of series from the above formula in the equation (vii) to get the answer.

$S=\dfrac{1}{4}\left[ \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+n+2\dfrac{n\left( n+1 \right)}{2} \right]$

Taking out ‘n’, we get

\[\begin{align}

& S=\dfrac{n}{4}\left[ \dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}+1+\left( n+1 \right) \right] \\

& S=\dfrac{n}{4}\left[ \dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}+n+2 \right] \\

\end{align}\]

Taking the LCM, we get

$\begin{align}

& S=\dfrac{n}{4}\left[ \dfrac{\left( n+1 \right)\left( 2n+1 \right)+6n+12}{6} \right] \\

& S=\dfrac{n}{24}\left[ 2{{n}^{2}}+2n+n+1+6n+12 \right] \\

& S=\dfrac{n\left( 2{{n}^{2}}+9n+13 \right)}{24} \\

\end{align}$

Hence, the sum of series given in the problem is $\dfrac{n\left( 2{{n}^{2}}+9n+13 \right)}{24}$.

Note: We can find the sum by taking ${{T}_{n}}=\dfrac{{{\left( n+1 \right)}^{2}}}{4}$itself without expanding it in following ways.

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

Now put (n+1) in place of ‘n’ in the formula of sum of series $\sum{{{n}^{2}}}$ i.e., $\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ .

Hence, we will get

$\dfrac{1}{4}\left( \dfrac{\left( n+1 \right)\left( n+2 \right)\left( 2n+3 \right)}{6}-1 \right)$

We will get the same answer as found in the solution.

Don’t confuse it with the sign of $\sum{{}}$(summation). It represents a series in its nth term (general term) in very short form. So, it’s nothing but a representation of any series in a short way. And be clear with the summation of the special series $\sum{n},\sum{{{n}^{2}},}\sum{{{n}^{3}}}$ . Without the summation of these series, we will not be able to solve these kinds of series.

Getting ${{T}_{n}}$ and hence applying $\sum{{}}$ (summation) to ${{T}_{n}}$is the key point of the question.

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