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Question Number 37991 by Fawomath last updated on 20/Jun/18

$$1.Findthesum$$$$s_n=1+2x+3x^2+4x^3+...+{nx}^{n-1}$$$$Hence,orotherwise,findthesum$$$$\underset{k=1}{\sum ^n}k.2^k$$$$2.Simplifythefollowing$$$$i.\underset{r=0}{\sum ^n}\left({}_{2r-1}^{2n}\right)$$$$ii.\underset{r=0}{\sum ^n}{(-1)}^{r}r\left({}_r^n\right)$$$$iii.\underset{r=0}{\sum ^n}{(-1)}^r\frac{1}{r+1}\left({}_r^n\right)$$$$iv.\underset{r=0}{\sum ^n}\left({}_{2r}^{2n}\right)$$$$v.\underset{r=0}{\sum ^n}{(-1)}^r\left({}_{n-r}^{n+1}\right)$$$$3.Findthesum$$$$\underset{r=0}{\sum ^{n-k}}\left({}_k^{n-r}\right),wherek=0,1,2,3,...,n$$

Commented by prof Abdo imad last updated on 21/Jun/18

$$1)letf\left(x\right)=1+x+x^2+...+x^n=\sum _{k=0}^n{x}^k$$$$\Rightarrow f^{{}^{\prime }}\left(x\right)=\sum _{k=1}^{n}k{x}^{k-1}butifx\ne 1$$$$f\left(x\right)=\frac{x^{n+1}-1}{x-1}\Rightarrow f^{{}^{\prime }}\left(x\right)=\frac{{nx}^{n+1}-(n+1)x^{n+1}+1}{{(x-1)}^2}$$$$so\sum _{k=1}^n{kx}^{k-1}=\frac{{nx}^{n+1}-(n+1)x^n+1}{{(x-1)}^2}$$$$ifx=1f^{{}^{\prime }}\left(x\right)=1+2+3+...+n$$$$\Rightarrow f^{{}^{\prime }}\left(x\right)=\frac{n(n+1)}{2}$$$$lettakex=2\Rightarrow \sum _{k=1}^{n}k.2^{k-1}=n{2}^{n+1}-(n+1)2^n+1$$$$\sum _{k=1}^{n}k.2^k=n{2}^{n+2}-(n+1)2^{n+1}+2.$$$$=4n{2}^n-(2n+2)2^n+2=(2n-2)2^n+2$$$$=(n-1)2^{n+1}+2.$$$$$$

Commented by prof Abdo imad last updated on 21/Jun/18

$$2)iiletp\left(x\right)=\sum _{k=0}^n{C}_n^k{x}^{k}wehave$$$$p\left(x\right)={(x+1)}^n\Rightarrow \sum _{k=0}^n{(-1)}^k{C}_n^k=p(-1)=0$$

Commented by prof Abdo imad last updated on 21/Jun/18

$$iiiletp\left(x\right)=\sum _{k=0}^n{C}_n^k{x}^k={(x+1)}^n\Rightarrow$$$$\int p\left(x\right)dx=\sum _{k=0}^n\frac{1}{k+1}{C}_n^k{x}^{k+1}+c=\frac{1}{n+1}{(x+1)}^{n+1}+c$$$$x=0\Rightarrow c=-\frac{1}{n+1}\Rightarrow$$$$\sum _{k=0}^n\frac{x^{k+1}}{k+1}{C}_n^k=\frac{{(x+1)}^{n+1}-1}{n+1}lettakex=-1\Rightarrow$$$$$$$$-\sum _{k=0}^n\frac{{(-1)}^k}{k+1}{C}_n^k=\frac{{(x+1)}^{n+1}-1}{n+1}\Rightarrow$$$$\sum _{k=0}^n\frac{{(-1)}^k}{k+1}{C}_n^k=\frac{1-{(x+1)}^{n+1}}{n+1}.$$$$$$

Commented by prof Abdo imad last updated on 21/Jun/18

$$\Rightarrow \sum _{k=0}^n\frac{{(-1)}^k}{k+1}{C}_n^k=\frac{1}{n+1}.$$

Answered by ajfour last updated on 20/Jun/18

$$s_n=1+2x+3x^2+4x^3+....+{nx}^{n-1}$$$${xs}_n=[x+2x^2+3x^3+....+(n-1)x^{n-1}+{nx}^n$$$$(1-x)s_n=1+x+x^2+...+x^{n-1}-{nx}^n$$$$s_n=\frac{1-x^n}{{(1-x)}^2}-\frac{{nx}^n}{1-x}.$$

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