Question Number 219078 by zetamaths last updated on 19/Apr/25
$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{m}}\right).\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \centerdot{k}\centerdot\frac{\left({m}!\right)^{\mathrm{2}} }{\left({m}−{k}\right)!\left({m}+{k}\right)!}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:{Proof}\:{this}\:{formula} \\ $$
Answered by MrGaster last updated on 19/Apr/25
![Let S(m)Σ_(k=1) ^m (−1)^(k−1) ∙k∙(((m!)^2 )/((m−k)!(m+k)!)) Denominator (m−k)!(m+k)! can bei assocated with combinationn umber (((2m)),((m+k)) )Rewritten as: S(m)=(((m!)^2 )/((2m)!))Σ_(k=1) ^m (−1)^(k−1) k (((2m)),((m+k)) ) let j=m+k Σ_(j=m+1) ^(2m) (−1)^(j−m−1) (j−m) (((2m)),(j) )= (((2m)),((2m−j)) ) [l=2m−j] Σ_(l=0) ^(m−1) (−1)^(m−l−1) (m−l) (((2m)),(l) ) The properties of alternationand entirety ofm cobination numbers are:Σ_(j=0) ^(2m) (−1)^j (((2m)),(j) )=0,Σ_(j=0) ^(2m) (−1)^j j (((2m)),(j) )=0. Split the summation ranged an substitute simplify to get: Σ_(k=1) ^m (−1)^(k−1) k (((2m)),((m+k)) )=((m∙(2m−1)!)/(2^(2m−1) (m−1)!)) Substitute and simplify thex epression Substitute them sumation result into S(m): S(m)=(((m!)^2 )/((2m)!))∙((m∙(2m−1)!)/(2^(2m−1) (m−1)!))=(m/(2(2m−1))) Combine the coefficients term to calculate the finalu reslt: The lefthand side of theoverall expression is: (1−(1/(2m)))S(m)=((2m−1)/(2m))∙(m/(2(2m−1))) PS:The equality is proven thet lefhand side is always equalo t the righthand side(1/4)](https://www.tinkutara.com/question/Q219080.png)
$$\mathrm{Let}\:{S}\left({m}\right)\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \centerdot{k}\centerdot\frac{\left({m}!\right)^{\mathrm{2}} }{\left({m}−{k}\right)!\left({m}+{k}\right)!} \\ $$$$\mathrm{Denominator}\:\left({m}−{k}\right)!\left({m}+{k}\right)!\:\mathrm{can}\:\mathrm{bei} \\ $$$$\mathrm{assocated}\:\mathrm{with}\:\mathrm{combinationn} \\ $$$$\mathrm{umber}\:\begin{pmatrix}{\mathrm{2}{m}}\\{{m}+{k}}\end{pmatrix}\mathrm{Rewritten}\:\mathrm{as}: \\ $$$${S}\left({m}\right)=\frac{\left({m}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{m}\right)!}\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {k}\begin{pmatrix}{\mathrm{2}{m}}\\{{m}+{k}}\end{pmatrix} \\ $$$$\mathrm{let}\:{j}={m}+{k} \underset{{j}={m}+\mathrm{1}} {\overset{\mathrm{2}{m}} {\sum}}\left(−\mathrm{1}\right)^{{j}−{m}−\mathrm{1}} \left({j}−{m}\right)\begin{pmatrix}{\mathrm{2}{m}}\\{{j}}\end{pmatrix}=\begin{pmatrix}{\mathrm{2}{m}}\\{\mathrm{2}{m}−{j}}\end{pmatrix} \\ $$$$\left[{l}=\mathrm{2}{m}−{j}\right] \underset{{l}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{m}−{l}−\mathrm{1}} \left({m}−{l}\right)\begin{pmatrix}{\mathrm{2}{m}}\\{{l}}\end{pmatrix} \\ $$$$\mathrm{The}\:\mathrm{properties}\:\mathrm{of}\: \\ $$$$\mathrm{alternationand}\:\mathrm{entirety}\:\mathrm{ofm} \\ $$$$\mathrm{cobination}\:\mathrm{numbers}\:\mathrm{are}:\underset{{j}=\mathrm{0}} {\overset{\mathrm{2}{m}} {\sum}}\left(−\mathrm{1}\right)^{{j}} \begin{pmatrix}{\mathrm{2}{m}}\\{{j}}\end{pmatrix}=\mathrm{0},\underset{{j}=\mathrm{0}} {\overset{\mathrm{2}{m}} {\sum}}\left(−\mathrm{1}\right)^{{j}} {j}\begin{pmatrix}{\mathrm{2}{m}}\\{{j}}\end{pmatrix}=\mathrm{0}. \\ $$$$\mathrm{Split}\:\mathrm{the}\:\mathrm{summation}\:\mathrm{ranged} \\ $$$$\mathrm{an}\:\mathrm{substitute}\:\mathrm{simplify}\:\mathrm{to}\:\mathrm{get}: \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {k}\begin{pmatrix}{\mathrm{2}{m}}\\{{m}+{k}}\end{pmatrix}=\frac{{m}\centerdot\left(\mathrm{2}{m}−\mathrm{1}\right)!}{\mathrm{2}^{\mathrm{2}{m}−\mathrm{1}} \left({m}−\mathrm{1}\right)!} \\ $$$$\mathrm{Substitute}\:\mathrm{and}\:\mathrm{simplify}\:\mathrm{thex} \\ $$$$\mathrm{epression}\:\:\:\mathrm{Substitute}\:\mathrm{them} \\ $$$$\mathrm{sumation}\:\mathrm{result}\:\mathrm{into}\:{S}\left({m}\right): \\ $$$${S}\left({m}\right)=\frac{\left({m}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{m}\right)!}\centerdot\frac{{m}\centerdot\left(\mathrm{2}{m}−\mathrm{1}\right)!}{\mathrm{2}^{\mathrm{2}{m}−\mathrm{1}} \left({m}−\mathrm{1}\right)!}=\frac{{m}}{\mathrm{2}\left(\mathrm{2}{m}−\mathrm{1}\right)} \\ $$$$\mathrm{Combine}\:\mathrm{the}\:\mathrm{coefficients} \\ $$$$\mathrm{term}\:\mathrm{to}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{finalu} \\ $$$$\mathrm{reslt}:\:\:\:\mathrm{The}\:\mathrm{lefthand}\:\mathrm{side}\:\mathrm{of}\: \\ $$$$\mathrm{theoverall}\:\mathrm{expression}\:\mathrm{is}: \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{m}}\right){S}\left({m}\right)=\frac{\mathrm{2}{m}−\mathrm{1}}{\mathrm{2}{m}}\centerdot\frac{{m}}{\mathrm{2}\left(\mathrm{2}{m}−\mathrm{1}\right)} \\ $$$$\mathrm{PS}:\mathrm{The}\:\mathrm{equality}\:\mathrm{is}\:\mathrm{proven}\:\mathrm{thet} \\ $$$$\mathrm{lefhand}\:\mathrm{side}\:\mathrm{is}\:\mathrm{always}\:\mathrm{equalo} \\ $$$$\mathrm{t}\:\mathrm{the}\:\mathrm{righthand}\:\mathrm{side}\frac{\mathrm{1}}{\mathrm{4}} \\ $$