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Question Number 208662 by efronzo1 last updated on 20/Jun/24

(( ((n),(0) ) +3 ((n),(1) ) +5 ((n),(2) ) +...+(2n+1) ((n),(n) ))/( ((n),(1) ) +2 ((n),(2) ) + 3 ((n),(3) ) +...+n ((n),(n) ))) =((23)/(11)) n=?

$$\:\:\frac{\begin{pmatrix}{\mathrm{n}}\\{\mathrm{0}}\end{pmatrix}\:+\mathrm{3}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{1}}\end{pmatrix}\:+\mathrm{5}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{2}}\end{pmatrix}\:+...+\left(\mathrm{2n}+\mathrm{1}\right)\begin{pmatrix}{\mathrm{n}}\\{\mathrm{n}}\end{pmatrix}}{\begin{pmatrix}{\mathrm{n}}\\{\mathrm{1}}\end{pmatrix}\:+\mathrm{2}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{2}}\end{pmatrix}\:+\:\mathrm{3}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{3}}\end{pmatrix}\:+...+\mathrm{n}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{n}}\end{pmatrix}}\:=\frac{\mathrm{23}}{\mathrm{11}} \\ $$$$\:\mathrm{n}=? \\ $$

Answered by Berbere last updated on 20/Jun/24

A=Σ_(k=0) ^n (2k+1) ((n),(k) );Σ_(k=0) ^n x^(2k+1) ((n),(k) )=f(x) =x(1+x^2 )^n ⇒A=f′(1) =2^n +2n.2^(n−1) =(n+1)2^n B=Σ_(k=0) ^n k ((n),(k) );g(x)=Σ ((n),(k) )x^k =(1+x)^n B=g′(1)=n.2^(n−1) ⇔(((n+1)2^n )/(n.2^(n−1) ))=((23)/(11))⇔((2(n+1))/n)=((23)/(11))⇒n=22

$${A}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{2}{k}+\mathrm{1}\right)\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix};\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{x}^{\mathrm{2}{k}+\mathrm{1}} \begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}={f}\left({x}\right) \\ $$$$={x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} \Rightarrow{A}={f}'\left(\mathrm{1}\right) \\ $$$$=\mathrm{2}^{{n}} +\mathrm{2}{n}.\mathrm{2}^{{n}−\mathrm{1}} =\left({n}+\mathrm{1}\right)\mathrm{2}^{{n}} \\ $$$${B}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{k}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix};{g}\left({x}\right)=\Sigma\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}{x}^{{k}} =\left(\mathrm{1}+{x}\right)^{{n}} \\ $$$${B}={g}'\left(\mathrm{1}\right)={n}.\mathrm{2}^{{n}−\mathrm{1}} \\ $$$$\Leftrightarrow\frac{\left({n}+\mathrm{1}\right)\mathrm{2}^{{n}} }{{n}.\mathrm{2}^{{n}−\mathrm{1}} }=\frac{\mathrm{23}}{\mathrm{11}}\Leftrightarrow\frac{\mathrm{2}\left({n}+\mathrm{1}\right)}{{n}}=\frac{\mathrm{23}}{\mathrm{11}}\Rightarrow{n}=\mathrm{22} \\ $$$$ \\ $$

Answered by Berbere last updated on 20/Jun/24

Σ_(k=1) ^n k ((n),(k) )=((k.n!)/(k!.(n−k)!))=nΣ_(k≥1) (((n−1)!)/((k−1)!.(n−1−(k−1))!)) =nΣ_(k=0) ^(n−1) ((n),(k) )=n2^(n−1) ..S Σ_(k=0) ^n (ak+b) ((n),(k) )=aS+b.2^n

$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}=\frac{{k}.{n}!}{{k}!.\left({n}−{k}\right)!}={n}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left({n}−\mathrm{1}\right)!}{\left({k}−\mathrm{1}\right)!.\left({n}−\mathrm{1}−\left({k}−\mathrm{1}\right)\right)!} \\ $$$$={n}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}={n}\mathrm{2}^{{n}−\mathrm{1}} ..{S} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({ak}+{b}\right)\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}={aS}+{b}.\mathrm{2}^{{n}} \\ $$

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