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Question Number 164544 by mr W last updated on 18/Jan/22

prove that Σ_(k=1) ^n ((n),(k) )^2 =(((2n)!)/((n!)^2 ))−1

$${prove}\:{that}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}^{\mathrm{2}} =\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} }−\mathrm{1} \\ $$

Answered by mindispower last updated on 18/Jan/22

=Σ_(k=0) ^n ((n),(k) )−1 =Σ_(k=0) ^n ((n),(k) ) ((n),((n−k)) ) Σ_(k=0) ^(2n) (((2n)),(k) )x^k =(1+x)^(2n) =(1+x)^n (1+x)^n =Σ_(k=0) ^n Σ_(j=0) ^n ((n),(k) ) ((n),((n−j)) )x^(n+k−j) coeficient of x^n =Σ_(k=0) ^n ((n),(k) ) ((n),((n−k)) )= (((2n)),(n) ) S= (((2n)),(n) )−1=(((2n)!)/((n!)^2 ))−1

$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}−\mathrm{1} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\begin{pmatrix}{{n}}\\{{n}−{k}}\end{pmatrix} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\begin{pmatrix}{\mathrm{2}{n}}\\{{k}}\end{pmatrix}{x}^{{k}} =\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} =\left(\mathrm{1}+{x}\right)^{{n}} \left(\mathrm{1}+{x}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\underset{{j}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\begin{pmatrix}{{n}}\\{{n}−{j}}\end{pmatrix}{x}^{{n}+{k}−{j}} \\ $$$${coeficient}\:{of}\:{x}^{{n}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\begin{pmatrix}{{n}}\\{{n}−{k}}\end{pmatrix}=\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix} \\ $$$${S}=\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}−\mathrm{1}=\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} }−\mathrm{1} \\ $$$$ \\ $$

Commented by mr W last updated on 18/Jan/22

thanks alot sir!

$${thanks}\:{alot}\:{sir}! \\ $$

Commented by mindispower last updated on 18/Jan/22

pleasur sir have nice day we can use integral also

$${pleasur}\:{sir}\:{have}\:\:{nice}\:{day}\:{we}\:{can}\:{use}\:{integral}\:{also} \\ $$

Answered by mindispower last updated on 19/Jan/22

∫_0 ^(2π) e^(imx) dx= { ((0 if m≠0)),((2π if m=0)) :} Σ_(k=0) ^n ((n),(k) )^2 =(1/(2π))Σ_(k=0) ^n Σ_(j=0) ^n ∫_0 ^(2π) ((n),(k) )e^(ikx) ((n),(j) )e^(−ijx) dx =(1/(2π))∫_0 ^(2π) (Σ_(k=0) ^n ((n),(k) )e^(ikx) )(Σ_(j=0) ^n ((n),(j) )e^(−ijx) )dx =(1/(2π))∫_0 ^(2π) (1+e^(ix) )^n (1+e^(−ix) )^n dx =(1/(2π))∫_0 ^(2π) (2+2cos(x))^n dx =(1/(2π))∫_0 ^(2π) 4^n cos^(2n) ((x/2))dx =(1/π) ∫_0 ^π 4^n coks^(2n) (y)dy=(1/π)(2∫_0 ^(π/2) cos^(2n) (y)dy) =(4^n /π)β(n+(1/2),(1/2)) =(1/π).((Γ(n+(1/2))Γ((1/2)))/(Γ(n+1)))=4^n (((n+(1/2)−1)......(n+(1/2)−(n−1))Γ^2 ((1/2)))/(n!.π)) =(2^n /(n!))Π_(k=0) ^(n−1) (2k+1) =((2^n .n!Π_(k=0) ^(n−1) (2k+1))/(n!.n!))=((Π_(k=1) ^n (2k).Π_(k=0) ^(n−1) (2k+1))/((n!)^2 ))=(((2n)!)/((n!)^2 )) Σ_(k=1) ^n ((n),(k) )^2 =Σ_(k=0) ^n ((n),(k) )^2 −1=(((2n)!)/((n!)^2 ))−1

$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} {e}^{{imx}} {dx}=\begin{cases}{\mathrm{0}\:{if}\:{m}\neq\mathrm{0}}\\{\mathrm{2}\pi\:{if}\:{m}=\mathrm{0}}\end{cases} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}\pi}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\underset{{j}=\mathrm{0}} {\overset{{n}} {\sum}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}{e}^{{ikx}} \begin{pmatrix}{{n}}\\{{j}}\end{pmatrix}{e}^{−{ijx}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left(\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}{e}^{{ikx}} \right)\left(\underset{{j}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{j}}\end{pmatrix}{e}^{−{ijx}} \right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left(\mathrm{1}+{e}^{{ix}} \right)^{{n}} \left(\mathrm{1}+{e}^{−{ix}} \right)^{{n}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left(\mathrm{2}+\mathrm{2}{cos}\left({x}\right)\right)^{{n}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{4}^{{n}} {cos}^{\mathrm{2}{n}} \left(\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\pi} \\ $$$$\int_{\mathrm{0}} ^{\pi} \mathrm{4}^{{n}} {coks}^{\mathrm{2}{n}} \left({y}\right){dy}=\frac{\mathrm{1}}{\pi}\left(\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}{n}} \left({y}\right){dy}\right) \\ $$$$=\frac{\mathrm{4}^{{n}} }{\pi}\beta\left({n}+\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\pi}.\frac{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}+\mathrm{1}\right)}=\mathrm{4}^{{n}} \frac{\left({n}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)......\left({n}+\frac{\mathrm{1}}{\mathrm{2}}−\left({n}−\mathrm{1}\right)\right)\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{{n}!.\pi} \\ $$$$=\frac{\mathrm{2}^{{n}} }{{n}!}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{2}{k}+\mathrm{1}\right) \\ $$$$=\frac{\mathrm{2}^{{n}} .{n}!\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{2}{k}+\mathrm{1}\right)}{{n}!.{n}!}=\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{2}{k}\right).\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{2}{k}+\mathrm{1}\right)}{\left({n}!\right)^{\mathrm{2}} }=\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} } \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}^{\mathrm{2}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}^{\mathrm{2}} −\mathrm{1}=\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} }−\mathrm{1} \\ $$$$ \\ $$

Commented by mr W last updated on 19/Jan/22

great!

$${great}! \\ $$

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