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Question: Prove that \(\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}\left( \begin{matrix} n \\\ k ...

Prove that ∑k=0n(−1)k(n k )1k+m+1=∑k=0m(−1)k(m k )1k+n+1\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}}

Explanation

Solution

Hint: Write expression of (1−x)n&(1−x)m{{\left( 1-x \right)}^{n}}\And {{\left( 1-x \right)}^{m}} and then apply integral to both sides.

Here, we have to prove
∑k=0n(−1)knCk1k+m+1=∑k=0m(−1)kmCk1k+n+1.........(1)\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=∑k=0n(−1)knck1(k+m+1)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
∑k=0n(−1)knck1(k+m+1)=nc0(m+1)−nc1(m+2)+nc2(m+3)+.......(−1)nncnm+n+1........(2)\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 nc0,nc1,nc2.......ncn−1,ncn{}^{n}{{c}_{0}},{}^{n}{{c}_{1}},{}^{n}{{c}_{2}}.......{}^{n}{{c}_{n-1}},{}^{n}{{c}_{n}} are coefficient of x0,x1,x2.......xn−1,xn in (1+x)n{{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;
(1+x)n=nc0+nc1x+nc2x2+.....ncnxn{{\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 (2)\left( 2 \right) has alternative positive and negative signs, means we need to relate the series by expansion of (1−x)n{{\left( 1-x \right)}^{n}} which can be written as
(1−x)n=nc0−nc1x+nc2x2−nc3x3+.........+(−1)nncnxn{{\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 xm{{x}^{m}} to both sides of the above series
(1−x)nxm=nc0xm−nc1xm+1+nc2xm+2+......(−1)nncnxn+m{{\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 ∫xn=xn+1n+1\int{{{x}^{n}}=\dfrac{{{x}^{n+1}}}{n+1}}
Let us integrate the above series from 0 to 10\text{ to 1} we get;
∫01(1−x)nxmdx=∫01nc0xmdx−∫01nc1xm+1dx+∫01nc2xm+2dx+.......(−1)n∫01ncnxm+ndx ∫01(1−x)nxmdx=nc0xm+1m+1∣1 0 −nc1xm+2m+2∣1 0 +nc2xm+3m+3∣1 0 +.......(−1)nncnxm+n+1m+n+1∣1 0  \begin{aligned} & \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{aligned}
Applying the limits, we get;
∫01(1−x)nxmdx=nc0m+1−nc1m+2+nc2m+3+........(−1)nncnm+n+1\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;
∑k=0n(−1)knck1k+m+1=∫01(1−x)nxmdx........(3)\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 (1−x)m{{\left( 1-x \right)}^{m}} as given summation can be expressed as:
∑k=0m(−1)kmck(k+n+1)=mc0n+1−mc1n+2+mc2n+3.........(−1)m−1mCm−1n+m+(−1)mmcmn+m+1\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}}
mc0,−mc1,mc2,−mc3...........{}^{m}{{c}_{0}},-{}^{m}{{c}_{1}},{}^{m}{{c}_{2}},-{}^{m}{{c}_{3}}........... are the coefficients of (1−x)m{{\left( 1-x \right)}^{m}} . Expansion of (1−x)m{{\left( 1-x \right)}^{m}}can be written as;
(1−x)m=mc0−mc1x+mc2x2−mc3x3+........(−1)m−1mcm−1xm−1+(−1)mmcmxm{{\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 xn{{x}^{n}} to both sides of the above expansion, we get
(1−x)mxn=mc0xn−mc1xn+1+mc2xn+2+.....(−1)m−1mcm−1xm+n−1+(−1)mmcmxm+n{{\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 to 10\text{ to }1
∫01(1−x)mxndx=∫01mc0xndx−∫01mc1xn+1dx+∫01mc2xn+2dx+.........∫01(−1)mmcmxm+ndx............(4)\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
∫xm=xm+1m+1\int{{{x}^{m}}=\dfrac{{{x}^{m+1}}}{m+1}}
Using the above formula in the equation (4)\left( 4 \right)
∫01(1−x)mxndx−mc0xn+1n+1∣1 0 −mc1xn+2n+2∣1 0 +........(−1)mmcmxn+m+1n+m+1∣1 0 \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
∫01(1−x)mxndx=mc0n+1−mc1n+2+mc2n+3+........(−1)mmcmn+m+1\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
∑k=0k=m(−1)kmckn+k+1=∫01(1−x)mxndx......(5)\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;
∫abf(x)dx=∫abf(a+b−x)dx\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 (5)\left( 5 \right) as
∑k=0k−m(−1)kmckn+k+1=∫01(1−x)mxndx=∫01(1−(0+1−x))m(0+1−x)ndx =∫01xm(1−x)ndx \begin{aligned} & \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{aligned}
Therefore,
∑k=0k=m(−1)kmckn+m+1=∫01xm(1−x)ndx............(6)\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 (3)&(6)\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.
∑k=0n(−1)knck1k+m+1=∑k=0m(−1)kmck1k+n+1\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 ∫01xm(1−x)ndx or ∫01xn(1−x)m\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 or n+1,n+2,....m+n+1m+1,m+2.....m+n+1\text{ or }n+1,n+2,....m+n+1 are in denominator and ∫xm=xm+1m+1\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 or n,n+1,n+2.....n+m+1m+1,m+2.....m+n+1\text{ or }n,n+1,n+2.....n+m+1 were in multiplication with the terms mc0,mc1,mc2...... or nc0,nc1,nc2.....ncn{}^{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 (1+x)n or (1+x)m{{\left( 1+x \right)}^{n}}\text{ or }{{\left( 1+x \right)}^{m}} and multiply it by xm or xn{{x}^{m}}\text{ or }{{x}^{n}} respectively, then integration the series from −1 to 0-1\text{ to }0 to get the required given series.