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

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

Answered by Berbere last updated on 20/Jun/24

$$A=\underset{k=0}{\sum ^n}(2k+1)\left(\begin{array}{c}n\\ k\end{array}\right);\underset{k=0}{\sum ^n}{x}^{2k+1}\left(\begin{array}{c}n\\ k\end{array}\right)=f\left(x\right)$$$$=x{(1+x^2)}^n\Rightarrow A=f^{\prime }\left(1\right)$$$$=2^n+2n.2^{n-1}=(n+1)2^n$$$$B=\underset{k=0}{\sum ^n}k\left(\begin{array}{c}n\\ k\end{array}\right);g\left(x\right)=\mathrm{\Sigma }\left(\begin{array}{c}n\\ k\end{array}\right)x^k={(1+x)}^n$$$$B=g^{\prime }\left(1\right)=n.2^{n-1}$$$$\iff \frac{(n+1)2^n}{n.2^{n-1}}=\frac{23}{11}\iff \frac{2(n+1)}{n}=\frac{23}{11}\Rightarrow n=22$$$$$$

Answered by Berbere last updated on 20/Jun/24

$$\underset{k=1}{\sum ^n}k\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{k.n!}{k!.(n-k)!}=n\sum _{k⩾1}\frac{(n-1)!}{(k-1)!.(n-1-(k-1))!}$$$$=n\underset{k=0}{\sum ^{n-1}}\left(\begin{array}{c}n\\ k\end{array}\right)=n{2}^{n-1}..S$$$$\underset{k=0}{\sum ^n}(ak+b)\left(\begin{array}{c}n\\ k\end{array}\right)=aS+b.2^n$$

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