Show that $\dfrac{1\times {{2}^{2}}+2\times {{3}^{3}}+...+n{{\left( n+1 \right)}^{2}}}{{{1}^{2}}\times 2+{{2}^{2}}\times 3+...+{{n}^{2}}\left( n+1 \right)}=\dfrac{3n+5}{3n+1}$.
Answer
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Hint:Here, we will separate the series of numerator and denominator and solve them individually. For solving series, we will convert it into summation series and simplify it and then we will use known answers of summation series to find our final answer. Common series that will be used in this question are –
(i) $\sum\limits_{i=1}^{n}{i}=n\left( \dfrac{n+1}{2} \right)$ that is sum of $1+2+3+4+5+...+n$
(ii) $\sum\limits_{i=1}^{n}{{{i}^{2}}}=n\dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}$ that is sum of ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+...+{{n}^{2}}$
(iii) $\sum\limits_{i=1}^{n}{{{i}^{3}}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}$ that is sum of ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+...+{{n}^{3}}$
Complete step-by-step solution
We are given the series $\dfrac{1\times {{2}^{2}}+2\times {{3}^{3}}+...+n{{\left( n+1 \right)}^{2}}}{{{1}^{2}}\times 2+{{2}^{2}}\times 3+...+{{n}^{2}}\left( n+1 \right)}$.
As we can see, ${{n}^{th}}$ term of the numerator $=n{{\left( n+1 \right)}^{2}}=n\left( {{n}^{2}}+1+2n \right)={{n}^{3}}+2{{n}^{2}}+n$
Also, ${{n}^{th}}$ term of the denominator $={{n}^{2}}\left( n+1 \right)={{n}^{3}}+{{n}^{2}}$.
Let us separate the numerator and the denominator series and solve them separately.
Hence, we can write series in the form as \[\sum\limits_{k=1}^{n}{{{k}^{3}}+}2{{k}^{2}}+k\]. Let us solve this series first. Separating summation for all terms, we get –
\[\sum\limits_{k=1}^{n}{{{k}^{3}}+}\sum\limits_{k=1}^{n}{2{{k}^{2}}}+\sum\limits_{k=1}^{n}{k}\]
Now as we know that,
$\begin{align}
&\Rightarrow \sum\limits_{i=1}^{n}{i}=n\left( \dfrac{n+1}{2} \right) \\
&\Rightarrow \sum\limits_{i=1}^{n}{{{i}^{2}}}=n\dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6} \\
&\Rightarrow \sum\limits_{i=1}^{n}{{{i}^{3}}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4} \\
\end{align}$
Using them for the given summation, we get –
$\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}+n\dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}+n\left( \dfrac{n+1}{2} \right)$
Taking $n\left( \dfrac{n+1}{2} \right)$ common, we get
\[n\dfrac{\left( n+1 \right)}{2}\left[ n\dfrac{\left( n+1 \right)}{2}+\dfrac{2\left( 2n+1 \right)}{3}+1 \right]\]
Taking LCM,
\[n\dfrac{\left( n+1 \right)}{2}\left[ \dfrac{3n\left( n+1 \right)+4\left( 2n+1 \right)+6}{6} \right]\]
Simplifying, we get –
\[\dfrac{n\left( n+1 \right)}{12}\left[ 3{{n}^{2}}+11n+10 \right]\]
\[\Rightarrow \dfrac{n\left( n+1 \right)}{12}\left[ 3{{n}^{2}}+6n+5n+10 \right]\]
Changing to factors, we get –
\[\dfrac{n\left( n+1 \right)\left( n+2 \right)\left( 3n+5 \right)}{12}...................(1)\]
Now, ${{n}^{th}}$ term of the denominator $={{n}^{3}}+{{n}^{2}}$.
Hence, we can write the series in the form as
\[\sum\limits_{k=1}^{n}{{{k}^{3}}+{{k}^{2}}}\]
Separating summation for all terms, we get –
\[\sum\limits_{k=1}^{n}{{{k}^{3}}+}\sum\limits_{k=1}^{n}{{{k}^{2}}}\]
As we know,
$\begin{align}
&\Rightarrow \sum\limits_{i=1}^{n}{{{i}^{2}}}=n\dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6} \\
&\Rightarrow \sum\limits_{i=1}^{n}{{{i}^{3}}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4} \\
\end{align}$
Using them, we get –
$\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}+n\dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}$
Taking $n\left( \dfrac{n+1}{2} \right)$ common, we get
\[n\dfrac{\left( n+1 \right)}{2}\left[ \dfrac{n\left( n+1 \right)}{2}+\dfrac{\left( 2n+1 \right)}{3} \right]\]
Taking LCM,
\[\begin{align}
& \dfrac{n\left( n+1 \right)}{2}\left[ \dfrac{3{{n}^{2}}+3n+4n+2}{6} \right] \\
&\Rightarrow \dfrac{n\left( n+1 \right)}{2}\left[ \dfrac{3{{n}^{2}}+7n+2}{6} \right] \\
\end{align}\]
Changing to factors, we get –
\[\begin{align}
& \dfrac{n\left( n+1 \right)}{12}\left[ 3n\left( n+2 \right)+1\left( n+2 \right) \right] \\
&\Rightarrow \dfrac{n\left( n+1 \right)\left( n+2 \right)\left( 3n+1 \right)}{12}.....................(2) \\
\end{align}\]
As we know, equation (1) represents summation of numerator of given series, equation (2) represents summation of denominator of given series.
Hence, putting them in given equation, we get –
$\dfrac{1\times {{2}^{2}}+2\times {{3}^{3}}+...+n{{\left( n+1 \right)}^{2}}}{{{1}^{2}}\times 2+{{2}^{2}}\times 3+...+{{n}^{2}}\left( n+1 \right)}=\dfrac{\dfrac{n\left( n+1 \right)\left( n+2 \right)\left( 3n+5 \right)}{12}}{\dfrac{n\left( n+1 \right)\left( n+2 \right)\left( 3n+1 \right)}{12}}$
$\begin{align}
& =\dfrac{n\left( n+1 \right)\left( n+2 \right)\left( 3n+5 \right)}{12}\times \dfrac{12}{n\left( n+1 \right)\left( n+2 \right)\left( 3n+1 \right)} \\
& =\left( \dfrac{3n+5}{3n+1} \right) \\
\end{align}$
Hence, the given result is proved.
Note: The students should know how to convert into summation series for solving these sums. They should remember basic summation formula such that \[\sum{n},{{\sum{n}}^{2}},{{\sum{n}}^{3}}\]. Calculations in this sum are quite difficult and students should do them carefully step by step. Converting the equation to factor is important so that we can cancel them out at last; otherwise, we will get stuck and cannot find the required answer.
(i) $\sum\limits_{i=1}^{n}{i}=n\left( \dfrac{n+1}{2} \right)$ that is sum of $1+2+3+4+5+...+n$
(ii) $\sum\limits_{i=1}^{n}{{{i}^{2}}}=n\dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}$ that is sum of ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+...+{{n}^{2}}$
(iii) $\sum\limits_{i=1}^{n}{{{i}^{3}}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}$ that is sum of ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+...+{{n}^{3}}$
Complete step-by-step solution
We are given the series $\dfrac{1\times {{2}^{2}}+2\times {{3}^{3}}+...+n{{\left( n+1 \right)}^{2}}}{{{1}^{2}}\times 2+{{2}^{2}}\times 3+...+{{n}^{2}}\left( n+1 \right)}$.
As we can see, ${{n}^{th}}$ term of the numerator $=n{{\left( n+1 \right)}^{2}}=n\left( {{n}^{2}}+1+2n \right)={{n}^{3}}+2{{n}^{2}}+n$
Also, ${{n}^{th}}$ term of the denominator $={{n}^{2}}\left( n+1 \right)={{n}^{3}}+{{n}^{2}}$.
Let us separate the numerator and the denominator series and solve them separately.
Hence, we can write series in the form as \[\sum\limits_{k=1}^{n}{{{k}^{3}}+}2{{k}^{2}}+k\]. Let us solve this series first. Separating summation for all terms, we get –
\[\sum\limits_{k=1}^{n}{{{k}^{3}}+}\sum\limits_{k=1}^{n}{2{{k}^{2}}}+\sum\limits_{k=1}^{n}{k}\]
Now as we know that,
$\begin{align}
&\Rightarrow \sum\limits_{i=1}^{n}{i}=n\left( \dfrac{n+1}{2} \right) \\
&\Rightarrow \sum\limits_{i=1}^{n}{{{i}^{2}}}=n\dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6} \\
&\Rightarrow \sum\limits_{i=1}^{n}{{{i}^{3}}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4} \\
\end{align}$
Using them for the given summation, we get –
$\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}+n\dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}+n\left( \dfrac{n+1}{2} \right)$
Taking $n\left( \dfrac{n+1}{2} \right)$ common, we get
\[n\dfrac{\left( n+1 \right)}{2}\left[ n\dfrac{\left( n+1 \right)}{2}+\dfrac{2\left( 2n+1 \right)}{3}+1 \right]\]
Taking LCM,
\[n\dfrac{\left( n+1 \right)}{2}\left[ \dfrac{3n\left( n+1 \right)+4\left( 2n+1 \right)+6}{6} \right]\]
Simplifying, we get –
\[\dfrac{n\left( n+1 \right)}{12}\left[ 3{{n}^{2}}+11n+10 \right]\]
\[\Rightarrow \dfrac{n\left( n+1 \right)}{12}\left[ 3{{n}^{2}}+6n+5n+10 \right]\]
Changing to factors, we get –
\[\dfrac{n\left( n+1 \right)\left( n+2 \right)\left( 3n+5 \right)}{12}...................(1)\]
Now, ${{n}^{th}}$ term of the denominator $={{n}^{3}}+{{n}^{2}}$.
Hence, we can write the series in the form as
\[\sum\limits_{k=1}^{n}{{{k}^{3}}+{{k}^{2}}}\]
Separating summation for all terms, we get –
\[\sum\limits_{k=1}^{n}{{{k}^{3}}+}\sum\limits_{k=1}^{n}{{{k}^{2}}}\]
As we know,
$\begin{align}
&\Rightarrow \sum\limits_{i=1}^{n}{{{i}^{2}}}=n\dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6} \\
&\Rightarrow \sum\limits_{i=1}^{n}{{{i}^{3}}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4} \\
\end{align}$
Using them, we get –
$\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}+n\dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}$
Taking $n\left( \dfrac{n+1}{2} \right)$ common, we get
\[n\dfrac{\left( n+1 \right)}{2}\left[ \dfrac{n\left( n+1 \right)}{2}+\dfrac{\left( 2n+1 \right)}{3} \right]\]
Taking LCM,
\[\begin{align}
& \dfrac{n\left( n+1 \right)}{2}\left[ \dfrac{3{{n}^{2}}+3n+4n+2}{6} \right] \\
&\Rightarrow \dfrac{n\left( n+1 \right)}{2}\left[ \dfrac{3{{n}^{2}}+7n+2}{6} \right] \\
\end{align}\]
Changing to factors, we get –
\[\begin{align}
& \dfrac{n\left( n+1 \right)}{12}\left[ 3n\left( n+2 \right)+1\left( n+2 \right) \right] \\
&\Rightarrow \dfrac{n\left( n+1 \right)\left( n+2 \right)\left( 3n+1 \right)}{12}.....................(2) \\
\end{align}\]
As we know, equation (1) represents summation of numerator of given series, equation (2) represents summation of denominator of given series.
Hence, putting them in given equation, we get –
$\dfrac{1\times {{2}^{2}}+2\times {{3}^{3}}+...+n{{\left( n+1 \right)}^{2}}}{{{1}^{2}}\times 2+{{2}^{2}}\times 3+...+{{n}^{2}}\left( n+1 \right)}=\dfrac{\dfrac{n\left( n+1 \right)\left( n+2 \right)\left( 3n+5 \right)}{12}}{\dfrac{n\left( n+1 \right)\left( n+2 \right)\left( 3n+1 \right)}{12}}$
$\begin{align}
& =\dfrac{n\left( n+1 \right)\left( n+2 \right)\left( 3n+5 \right)}{12}\times \dfrac{12}{n\left( n+1 \right)\left( n+2 \right)\left( 3n+1 \right)} \\
& =\left( \dfrac{3n+5}{3n+1} \right) \\
\end{align}$
Hence, the given result is proved.
Note: The students should know how to convert into summation series for solving these sums. They should remember basic summation formula such that \[\sum{n},{{\sum{n}}^{2}},{{\sum{n}}^{3}}\]. Calculations in this sum are quite difficult and students should do them carefully step by step. Converting the equation to factor is important so that we can cancel them out at last; otherwise, we will get stuck and cannot find the required answer.
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