Prove that $\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}\left( \begin{matrix} n \\ k \\\end{matrix} \right)}\dfrac{1}{k+m+1}=\sum\limits_{k=0}^{m}{{{\left( -1 \right)}^{k}}\left( \begin{matrix} m \\ k \\\end{matrix} \right)\dfrac{1}{k+n+1}}$

Question

Prove that $\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}\left( \begin{matrix}   n  \\   k  \\\end{matrix} \right)}\dfrac{1}{k+m+1}=\sum\limits_{k=0}^{m}{{{\left( -1 \right)}^{k}}\left( \begin{matrix}   m  \\   k  \\\end{matrix} \right)\dfrac{1}{k+n+1}}$

Vedantu Content Team · Accepted Answer

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.

Prove that ∑k=0n(−1)k(nk)1k+m+1=∑k=0m(−1)k(mk)1k+n+1

Prove that $\sum_{k = 0}^{n} {(- 1)}^{k} (\begin{matrix} n \\ k \end{matrix}) \frac{1}{k + m + 1} = \sum_{k = 0}^{m} {(- 1)}^{k} (\begin{matrix} m \\ k \end{matrix}) \frac{1}{k + n + 1}$