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Prove that $\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}\left( \begin{matrix}
n \\
k \\
\end{matrix} \right)}\dfrac{1}{k+m+1}=\sum\limits_{k=0}^{m}{{{\left( -1 \right)}^{k}}\left( \begin{matrix}
m \\
k \\
\end{matrix} \right)\dfrac{1}{k+n+1}}$
Answer
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Hint: Write expression of ${{\left( 1-x \right)}^{n}}\And {{\left( 1-x \right)}^{m}}$ and then apply integral to both sides.
Here, we have to prove
$\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}{}^{n}{{C}_{k}}}\dfrac{1}{k+m+1}=\sum\limits_{k=0}^{m}{{{\left( -1 \right)}^{k}}{}^{m}{{C}_{k}}\dfrac{1}{k+n+1}}.........\left( 1 \right)$
Now, we cannot concert LHS to RHS directly, so basically we need to simplify LHS and RHS both for proving.
Let us simplifying LHS part:
\[LHS=\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}{}^{n}{{c}_{k}}\dfrac{1}{\left( k+m+1 \right)}}\]
Writing the above summation to series as
\[\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}{}^{n}{{c}_{k}}\dfrac{1}{\left( k+m+1 \right)}=\dfrac{{}^{n}{{c}_{0}}}{\left( m+1 \right)}-\dfrac{{}^{n}{{c}_{1}}}{\left( m+2 \right)}+\dfrac{{}^{n}{{c}_{2}}}{\left( m+3 \right)}}+.......{{\left( -1 \right)}^{n}}\dfrac{{}^{n}{{c}_{n}}}{m+n+1}........\left( 2 \right)\]
Now we can observe that ${}^{n}{{c}_{0}},{}^{n}{{c}_{1}},{}^{n}{{c}_{2}}.......{}^{n}{{c}_{n-1}},{}^{n}{{c}_{n}}$ are coefficient of ${{x}^{0}},{{x}^{1}},{{x}^{2}}.......{{x}^{n-1}},{{x}^{n}}\text{ in }{{\left( 1+x \right)}^{n}}$ as expansion of it can be written as;
${{\left( 1+x \right)}^{n}}={}^{n}{{c}_{0}}+{}^{n}{{c}_{1}}x+{}^{n}{{c}_{2}}{{x}^{2}}+.....{}^{n}{{c}_{n}}{{x}^{n}}$
As, series of equation $\left( 2 \right)$ has alternative positive and negative signs, means we need to relate the series by expansion of ${{\left( 1-x \right)}^{n}}$ which can be written as
\[{{\left( 1-x \right)}^{n}}={}^{n}{{c}_{0}}-{}^{n}{{c}_{1}}x+{}^{n}{{c}_{2}}{{x}^{2}}-{}^{n}{{c}_{3}}{{x}^{3}}+.........+{{\left( -1 \right)}^{n}}{}^{n}{{c}_{n}}{{x}^{n}}\]
Let us multiply by ${{x}^{m}}$ to both sides of the above series
${{\left( 1-x \right)}^{n}}{{x}^{m}}={}^{n}{{c}_{0}}{{x}^{m}}-{}^{n}{{c}_{1}}{{x}^{m+1}}+{}^{n}{{c}_{2}}{{x}^{m+2}}+......{{\left( -1 \right)}^{n}}{}^{n}{{c}_{n}}{{x}^{n+m}}$
We have \[\int{{{x}^{n}}=\dfrac{{{x}^{n+1}}}{n+1}}\]
Let us integrate the above series from $0\text{ to 1}$ we get;
$\begin{align}
& \int_{0}^{1}{{{\left( 1-x \right)}^{n}}{{x}^{m}}dx}=\int_{0}^{1}{{}^{n}{{c}_{0}}{{x}^{m}}dx}-\int_{0}^{1}{{}^{n}{{c}_{1}}{{x}^{m+1}}dx}+\int_{0}^{1}{{}^{n}{{c}_{2}}{{x}^{m+2}}dx}+.......{{\left( -1 \right)}^{n}}\int_{0}^{1}{{}^{n}{{c}_{n}}{{x}^{m+n}}dx} \\
& \int_{0}^{1}{{{\left( 1-x \right)}^{n}}{{x}^{m}}dx}={}^{n}{{c}_{0}}\dfrac{{{x}^{m+1}}}{m+1}\left| \begin{matrix}
1 \\
0 \\
\end{matrix} \right.-{}^{n}{{c}_{1}}\dfrac{{{x}^{m+2}}}{m+2}\left| \begin{matrix}
1 \\
0 \\
\end{matrix} \right.+{}^{n}{{c}_{2}}\dfrac{{{x}^{m+3}}}{m+3}\left| \begin{matrix}
1 \\
0 \\
\end{matrix} \right.+.......{{\left( -1 \right)}^{n}}{}^{n}{{c}_{n}}\dfrac{{{x}^{m+n+1}}}{m+n+1}\left| \begin{matrix}
1 \\
0 \\
\end{matrix} \right. \\
\end{align}$
Applying the limits, we get;
$\int_{0}^{1}{{{\left( 1-x \right)}^{n}}{{x}^{m}}dx}=\dfrac{{}^{n}{{c}_{0}}}{m+1}-\dfrac{{}^{n}{{c}_{1}}}{m+2}+\dfrac{{}^{n}{{c}_{2}}}{m+3}+........{{\left( -1 \right)}^{n}}\dfrac{{}^{n}{{c}_{n}}}{m+n+1}$
Hence, LHS part can be written as;
$\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}{}^{n}{{c}_{k}}\dfrac{1}{k+m+1}=\int_{0}^{1}{{{\left( 1-x \right)}^{n}}{{x}^{m}}dx........\left( 3 \right)}}$
Now, let us simplify the RHS part in a similar way. Here we have to take expansion of ${{\left( 1-x \right)}^{m}}$ as given summation can be expressed as:
\[\sum\limits_{k=0}^{m}{{{\left( -1 \right)}^{k}}\dfrac{{}^{m}{{c}_{k}}}{\left( k+n+1 \right)}=\dfrac{{}^{m}{{c}_{0}}}{n+1}-\dfrac{{}^{m}{{c}_{1}}}{n+2}+\dfrac{{}^{m}{{c}_{2}}}{n+3}.........{{\left( -1 \right)}^{m-1}}\dfrac{{}^{m}{{C}_{m-1}}}{n+m}+{{\left( -1 \right)}^{m}}\dfrac{{}^{m}{{c}_{m}}}{n+m+1}}\]
${}^{m}{{c}_{0}},-{}^{m}{{c}_{1}},{}^{m}{{c}_{2}},-{}^{m}{{c}_{3}}...........$ are the coefficients of ${{\left( 1-x \right)}^{m}}$ . Expansion of ${{\left( 1-x \right)}^{m}}$can be written as;
${{\left( 1-x \right)}^{m}}={}^{m}{{c}_{0}}-{}^{m}{{c}_{1}}x+{}^{m}{{c}_{2}}{{x}^{2}}-{}^{m}{{c}_{3}}{{x}^{3}}+........{{\left( -1 \right)}^{m-1}}{}^{m}{{c}_{m-1}}{{x}^{m-1}}+{{\left( -1 \right)}^{m}}{}^{m}{{c}_{m}}{{x}^{m}}$
Multiplying by ${{x}^{n}}$ to both sides of the above expansion, we get
${{\left( 1-x \right)}^{m}}{{x}^{n}}={}^{m}{{c}_{0}}{{x}^{n}}-{}^{m}{{c}_{1}}{{x}^{n+1}}+{}^{m}{{c}_{2}}{{x}^{n+2}}+.....{{\left( -1 \right)}^{m-1}}{}^{m}{{c}_{m-1}}{{x}^{m+n-1}}+{{\left( -1 \right)}^{m}}{}^{m}{{c}_{m}}{{x}^{m+n}}$
Integrating the above series to both sides from the limit $0\text{ to }1$
$\int\limits_{0}^{1}{{{\left( 1-x \right)}^{m}}{{x}^{n}}dx}=\int_{0}^{1}{{}^{m}{{c}_{0}}{{x}^{n}}dx-}\int_{0}^{1}{{}^{m}{{c}_{1}}{{x}^{n+1}}dx+\int_{0}^{1}{{}^{m}{{c}_{2}}{{x}^{n+2}}dx+.........\int_{0}^{1}{{{\left( -1 \right)}^{m}}{}^{m}{{c}_{m}}{{x}^{m+n}}dx............\left( 4 \right)}}}$
We have
$\int{{{x}^{m}}=\dfrac{{{x}^{m+1}}}{m+1}}$
Using the above formula in the equation $\left( 4 \right)$
$\int_{0}^{1}{{{\left( 1-x \right)}^{m}}{{x}^{n}}dx}-\dfrac{{}^{m}{{c}_{0}}{{x}^{n+1}}}{n+1}\left| \begin{matrix}
1 \\
0 \\
\end{matrix}- \right.\dfrac{{}^{m}{{c}_{1}}{{x}^{n+2}}}{n+2}\left| \begin{matrix}
1 \\
0 \\
\end{matrix} \right.+........{{\left( -1 \right)}^{m}}\dfrac{{}^{m}{{c}_{m}}{{x}^{n+m+1}}}{n+m+1}\left| \begin{matrix}
1 \\
0 \\
\end{matrix} \right.$
Applying the limits, we get
$\int_{0}^{1}{{{\left( 1-x \right)}^{m}}{{x}^{n}}dx}=\dfrac{{}^{m}{{c}_{0}}}{n+1}-\dfrac{{}^{m}{{c}_{1}}}{n+2}+\dfrac{{}^{m}{{c}_{2}}}{n+3}+........{{\left( -1 \right)}^{m}}\dfrac{{}^{m}{{c}_{m}}}{n+m+1}$
Rewriting the above equation in summation form we will get RHS part as
$\sum\limits_{k=0}^{k=m}{{{\left( -1 \right)}^{k}}\dfrac{{}^{m}{{c}_{k}}}{n+k+1}}=\int_{0}^{1}{{{\left( 1-x \right)}^{m}}{{x}^{n}}dx}......\left( 5 \right)$
We have a property of definite integral as;
$\int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{b}{f\left( a+b-x \right)dx}$
We can use the above property with equation $\left( 5 \right)$ as
$\begin{align}
& \sum\limits_{k=0}^{k-m}{\dfrac{{{\left( -1 \right)}^{k}}{}^{m}{{c}_{k}}}{n+k+1}=\int_{0}^{1}{{{\left( 1-x \right)}^{m}}{{x}^{n}}dx}=\int_{0}^{1}{{{\left( 1-\left( 0+1-x \right) \right)}^{m}}{{\left( 0+1-x \right)}^{n}}dx}} \\
& =\int_{0}^{1}{{{x}^{m}}{{\left( 1-x \right)}^{n}}dx} \\
\end{align}$
Therefore,
$\sum\limits_{k=0}^{k=m}{\dfrac{{{\left( -1 \right)}^{k}}{}^{m}{{c}_{k}}}{n+m+1}=\int_{0}^{1}{{{x}^{m}}{{\left( 1-x \right)}^{n}}dx............\left( 6 \right)}}$
Now, comparing the equation $\left( 3 \right)\And \left( 6 \right)$ ,the RHS part of both the equations are equal, hence, the LHS part of the equation should also be equal.
$\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}{}^{n}{{c}_{k}}\dfrac{1}{k+m+1}=}\sum\limits_{k=0}^{m}{{{\left( -1 \right)}^{k}}{}^{m}{{c}_{k}}\dfrac{1}{k+n+1}}$
Hence proved.
Note: No need to solve $\int_{0}^{1}{{{x}^{m}}{{\left( 1-x \right)}^{n}}dx\text{ or }}\int_{0}^{1}{{{x}^{n}}{{\left( 1-x \right)}^{m}}}$ further. As we can use the property of definite integral. One can waste his/her time with the integral part.
One can think why integration is used, the reason is simple terms $m+1,m+2.....m+n+1\text{ or }n+1,n+2,....m+n+1$ are in denominator and $\int{{{x}^{m}}=\dfrac{{{x}^{m+1}}}{m+1}}$ , hence we need use integration only by observation of the given series. If the terms $m+1,m+2.....m+n+1\text{ or }n,n+1,n+2.....n+m+1$ were in multiplication with the terms ${}^{m}{{c}_{0}},{}^{m}{{c}_{1}},{}^{m}{{c}_{2}}......\text{ or }{}^{n}{{c}_{0}},{}^{n}{{c}_{1}},{}^{n}{{c}_{2}}.....{}^{n}{{c}_{n}}$ then we need to use concept of differentiation. Hence observation is the key point of this question.
Another approach would be that we can take expansion of ${{\left( 1+x \right)}^{n}}\text{ or }{{\left( 1+x \right)}^{m}}$ and multiply it by ${{x}^{m}}\text{ or }{{x}^{n}}$ respectively, then integration the series from $-1\text{ to }0$ to get the required given series.
Here, we have to prove
$\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}{}^{n}{{C}_{k}}}\dfrac{1}{k+m+1}=\sum\limits_{k=0}^{m}{{{\left( -1 \right)}^{k}}{}^{m}{{C}_{k}}\dfrac{1}{k+n+1}}.........\left( 1 \right)$
Now, we cannot concert LHS to RHS directly, so basically we need to simplify LHS and RHS both for proving.
Let us simplifying LHS part:
\[LHS=\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}{}^{n}{{c}_{k}}\dfrac{1}{\left( k+m+1 \right)}}\]
Writing the above summation to series as
\[\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}{}^{n}{{c}_{k}}\dfrac{1}{\left( k+m+1 \right)}=\dfrac{{}^{n}{{c}_{0}}}{\left( m+1 \right)}-\dfrac{{}^{n}{{c}_{1}}}{\left( m+2 \right)}+\dfrac{{}^{n}{{c}_{2}}}{\left( m+3 \right)}}+.......{{\left( -1 \right)}^{n}}\dfrac{{}^{n}{{c}_{n}}}{m+n+1}........\left( 2 \right)\]
Now we can observe that ${}^{n}{{c}_{0}},{}^{n}{{c}_{1}},{}^{n}{{c}_{2}}.......{}^{n}{{c}_{n-1}},{}^{n}{{c}_{n}}$ are coefficient of ${{x}^{0}},{{x}^{1}},{{x}^{2}}.......{{x}^{n-1}},{{x}^{n}}\text{ in }{{\left( 1+x \right)}^{n}}$ as expansion of it can be written as;
${{\left( 1+x \right)}^{n}}={}^{n}{{c}_{0}}+{}^{n}{{c}_{1}}x+{}^{n}{{c}_{2}}{{x}^{2}}+.....{}^{n}{{c}_{n}}{{x}^{n}}$
As, series of equation $\left( 2 \right)$ has alternative positive and negative signs, means we need to relate the series by expansion of ${{\left( 1-x \right)}^{n}}$ which can be written as
\[{{\left( 1-x \right)}^{n}}={}^{n}{{c}_{0}}-{}^{n}{{c}_{1}}x+{}^{n}{{c}_{2}}{{x}^{2}}-{}^{n}{{c}_{3}}{{x}^{3}}+.........+{{\left( -1 \right)}^{n}}{}^{n}{{c}_{n}}{{x}^{n}}\]
Let us multiply by ${{x}^{m}}$ to both sides of the above series
${{\left( 1-x \right)}^{n}}{{x}^{m}}={}^{n}{{c}_{0}}{{x}^{m}}-{}^{n}{{c}_{1}}{{x}^{m+1}}+{}^{n}{{c}_{2}}{{x}^{m+2}}+......{{\left( -1 \right)}^{n}}{}^{n}{{c}_{n}}{{x}^{n+m}}$
We have \[\int{{{x}^{n}}=\dfrac{{{x}^{n+1}}}{n+1}}\]
Let us integrate the above series from $0\text{ to 1}$ we get;
$\begin{align}
& \int_{0}^{1}{{{\left( 1-x \right)}^{n}}{{x}^{m}}dx}=\int_{0}^{1}{{}^{n}{{c}_{0}}{{x}^{m}}dx}-\int_{0}^{1}{{}^{n}{{c}_{1}}{{x}^{m+1}}dx}+\int_{0}^{1}{{}^{n}{{c}_{2}}{{x}^{m+2}}dx}+.......{{\left( -1 \right)}^{n}}\int_{0}^{1}{{}^{n}{{c}_{n}}{{x}^{m+n}}dx} \\
& \int_{0}^{1}{{{\left( 1-x \right)}^{n}}{{x}^{m}}dx}={}^{n}{{c}_{0}}\dfrac{{{x}^{m+1}}}{m+1}\left| \begin{matrix}
1 \\
0 \\
\end{matrix} \right.-{}^{n}{{c}_{1}}\dfrac{{{x}^{m+2}}}{m+2}\left| \begin{matrix}
1 \\
0 \\
\end{matrix} \right.+{}^{n}{{c}_{2}}\dfrac{{{x}^{m+3}}}{m+3}\left| \begin{matrix}
1 \\
0 \\
\end{matrix} \right.+.......{{\left( -1 \right)}^{n}}{}^{n}{{c}_{n}}\dfrac{{{x}^{m+n+1}}}{m+n+1}\left| \begin{matrix}
1 \\
0 \\
\end{matrix} \right. \\
\end{align}$
Applying the limits, we get;
$\int_{0}^{1}{{{\left( 1-x \right)}^{n}}{{x}^{m}}dx}=\dfrac{{}^{n}{{c}_{0}}}{m+1}-\dfrac{{}^{n}{{c}_{1}}}{m+2}+\dfrac{{}^{n}{{c}_{2}}}{m+3}+........{{\left( -1 \right)}^{n}}\dfrac{{}^{n}{{c}_{n}}}{m+n+1}$
Hence, LHS part can be written as;
$\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}{}^{n}{{c}_{k}}\dfrac{1}{k+m+1}=\int_{0}^{1}{{{\left( 1-x \right)}^{n}}{{x}^{m}}dx........\left( 3 \right)}}$
Now, let us simplify the RHS part in a similar way. Here we have to take expansion of ${{\left( 1-x \right)}^{m}}$ as given summation can be expressed as:
\[\sum\limits_{k=0}^{m}{{{\left( -1 \right)}^{k}}\dfrac{{}^{m}{{c}_{k}}}{\left( k+n+1 \right)}=\dfrac{{}^{m}{{c}_{0}}}{n+1}-\dfrac{{}^{m}{{c}_{1}}}{n+2}+\dfrac{{}^{m}{{c}_{2}}}{n+3}.........{{\left( -1 \right)}^{m-1}}\dfrac{{}^{m}{{C}_{m-1}}}{n+m}+{{\left( -1 \right)}^{m}}\dfrac{{}^{m}{{c}_{m}}}{n+m+1}}\]
${}^{m}{{c}_{0}},-{}^{m}{{c}_{1}},{}^{m}{{c}_{2}},-{}^{m}{{c}_{3}}...........$ are the coefficients of ${{\left( 1-x \right)}^{m}}$ . Expansion of ${{\left( 1-x \right)}^{m}}$can be written as;
${{\left( 1-x \right)}^{m}}={}^{m}{{c}_{0}}-{}^{m}{{c}_{1}}x+{}^{m}{{c}_{2}}{{x}^{2}}-{}^{m}{{c}_{3}}{{x}^{3}}+........{{\left( -1 \right)}^{m-1}}{}^{m}{{c}_{m-1}}{{x}^{m-1}}+{{\left( -1 \right)}^{m}}{}^{m}{{c}_{m}}{{x}^{m}}$
Multiplying by ${{x}^{n}}$ to both sides of the above expansion, we get
${{\left( 1-x \right)}^{m}}{{x}^{n}}={}^{m}{{c}_{0}}{{x}^{n}}-{}^{m}{{c}_{1}}{{x}^{n+1}}+{}^{m}{{c}_{2}}{{x}^{n+2}}+.....{{\left( -1 \right)}^{m-1}}{}^{m}{{c}_{m-1}}{{x}^{m+n-1}}+{{\left( -1 \right)}^{m}}{}^{m}{{c}_{m}}{{x}^{m+n}}$
Integrating the above series to both sides from the limit $0\text{ to }1$
$\int\limits_{0}^{1}{{{\left( 1-x \right)}^{m}}{{x}^{n}}dx}=\int_{0}^{1}{{}^{m}{{c}_{0}}{{x}^{n}}dx-}\int_{0}^{1}{{}^{m}{{c}_{1}}{{x}^{n+1}}dx+\int_{0}^{1}{{}^{m}{{c}_{2}}{{x}^{n+2}}dx+.........\int_{0}^{1}{{{\left( -1 \right)}^{m}}{}^{m}{{c}_{m}}{{x}^{m+n}}dx............\left( 4 \right)}}}$
We have
$\int{{{x}^{m}}=\dfrac{{{x}^{m+1}}}{m+1}}$
Using the above formula in the equation $\left( 4 \right)$
$\int_{0}^{1}{{{\left( 1-x \right)}^{m}}{{x}^{n}}dx}-\dfrac{{}^{m}{{c}_{0}}{{x}^{n+1}}}{n+1}\left| \begin{matrix}
1 \\
0 \\
\end{matrix}- \right.\dfrac{{}^{m}{{c}_{1}}{{x}^{n+2}}}{n+2}\left| \begin{matrix}
1 \\
0 \\
\end{matrix} \right.+........{{\left( -1 \right)}^{m}}\dfrac{{}^{m}{{c}_{m}}{{x}^{n+m+1}}}{n+m+1}\left| \begin{matrix}
1 \\
0 \\
\end{matrix} \right.$
Applying the limits, we get
$\int_{0}^{1}{{{\left( 1-x \right)}^{m}}{{x}^{n}}dx}=\dfrac{{}^{m}{{c}_{0}}}{n+1}-\dfrac{{}^{m}{{c}_{1}}}{n+2}+\dfrac{{}^{m}{{c}_{2}}}{n+3}+........{{\left( -1 \right)}^{m}}\dfrac{{}^{m}{{c}_{m}}}{n+m+1}$
Rewriting the above equation in summation form we will get RHS part as
$\sum\limits_{k=0}^{k=m}{{{\left( -1 \right)}^{k}}\dfrac{{}^{m}{{c}_{k}}}{n+k+1}}=\int_{0}^{1}{{{\left( 1-x \right)}^{m}}{{x}^{n}}dx}......\left( 5 \right)$
We have a property of definite integral as;
$\int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{b}{f\left( a+b-x \right)dx}$
We can use the above property with equation $\left( 5 \right)$ as
$\begin{align}
& \sum\limits_{k=0}^{k-m}{\dfrac{{{\left( -1 \right)}^{k}}{}^{m}{{c}_{k}}}{n+k+1}=\int_{0}^{1}{{{\left( 1-x \right)}^{m}}{{x}^{n}}dx}=\int_{0}^{1}{{{\left( 1-\left( 0+1-x \right) \right)}^{m}}{{\left( 0+1-x \right)}^{n}}dx}} \\
& =\int_{0}^{1}{{{x}^{m}}{{\left( 1-x \right)}^{n}}dx} \\
\end{align}$
Therefore,
$\sum\limits_{k=0}^{k=m}{\dfrac{{{\left( -1 \right)}^{k}}{}^{m}{{c}_{k}}}{n+m+1}=\int_{0}^{1}{{{x}^{m}}{{\left( 1-x \right)}^{n}}dx............\left( 6 \right)}}$
Now, comparing the equation $\left( 3 \right)\And \left( 6 \right)$ ,the RHS part of both the equations are equal, hence, the LHS part of the equation should also be equal.
$\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}{}^{n}{{c}_{k}}\dfrac{1}{k+m+1}=}\sum\limits_{k=0}^{m}{{{\left( -1 \right)}^{k}}{}^{m}{{c}_{k}}\dfrac{1}{k+n+1}}$
Hence proved.
Note: No need to solve $\int_{0}^{1}{{{x}^{m}}{{\left( 1-x \right)}^{n}}dx\text{ or }}\int_{0}^{1}{{{x}^{n}}{{\left( 1-x \right)}^{m}}}$ further. As we can use the property of definite integral. One can waste his/her time with the integral part.
One can think why integration is used, the reason is simple terms $m+1,m+2.....m+n+1\text{ or }n+1,n+2,....m+n+1$ are in denominator and $\int{{{x}^{m}}=\dfrac{{{x}^{m+1}}}{m+1}}$ , hence we need use integration only by observation of the given series. If the terms $m+1,m+2.....m+n+1\text{ or }n,n+1,n+2.....n+m+1$ were in multiplication with the terms ${}^{m}{{c}_{0}},{}^{m}{{c}_{1}},{}^{m}{{c}_{2}}......\text{ or }{}^{n}{{c}_{0}},{}^{n}{{c}_{1}},{}^{n}{{c}_{2}}.....{}^{n}{{c}_{n}}$ then we need to use concept of differentiation. Hence observation is the key point of this question.
Another approach would be that we can take expansion of ${{\left( 1+x \right)}^{n}}\text{ or }{{\left( 1+x \right)}^{m}}$ and multiply it by ${{x}^{m}}\text{ or }{{x}^{n}}$ respectively, then integration the series from $-1\text{ to }0$ to get the required given series.
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