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Question Number 8156 by 314159 last updated on 02/Oct/16

$$\mathrm{The}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{series}$$$$3{}^n{C}_0-8{}^n{C}_1+13{}^n{C}_2-18{}^n{C}_3+...+{(-1)}^n(3+5n){}^n{C}_n]$$

Commented by Yozzias last updated on 02/Oct/16

$$\underset{0}{\sum ^{\mathrm{n}}}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{r}\end{array}\right){(-1)}^{\mathrm{r}}(3+5\mathrm{r})=3\underset{0}{\sum ^{\mathrm{n}}}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{r}\end{array}\right){(-1)}^{\mathrm{r}}+5\underset{0}{\sum ^{\mathrm{n}}}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{r}\end{array}\right){(-1)}^{\mathrm{r}}\mathrm{r}$$$$\mathrm{Let}{(1-\mathrm{x})}^{\mathrm{n}}=\underset{0}{\sum ^{\mathrm{n}}}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{r}\end{array}\right){(-1)}^{\mathrm{r}}{\mathrm{x}}^{\mathrm{r}}.....\left(1\right)$$$$\mathrm{Differentiating}\Rightarrow -\mathrm{n}{(1-\mathrm{x})}^{\mathrm{n}-1}=\underset{0}{\sum ^{\mathrm{n}}}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{r}\end{array}\right)\mathrm{r}{(-1)}^{\mathrm{r}}{\mathrm{x}}^{\mathrm{r}-1}$$$$\mathrm{x}=1\Rightarrow 0=\underset{0}{\sum ^{\mathrm{n}}}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{r}\end{array}\right){(-1)}^{\mathrm{r}}\mathrm{r}.......\left(2\right)$$$$\mathrm{Also},\mathrm{x}=1\mathrm{in}\left(1\right)\Rightarrow \underset{0}{\sum ^{\mathrm{n}}}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{r}\end{array}\right){(-1)}^{\mathrm{r}}=0.$$$$\therefore \underset{0}{\sum ^{\mathrm{n}}}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{r}\end{array}\right){(-1)}^{\mathrm{r}}(3+5\mathrm{r})=3\times 0+0$$$$\underset{0}{\sum ^{\mathrm{n}}}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{r}\end{array}\right){(-1)}^{\mathrm{r}}(3+5\mathrm{r})=0$$$$$$

Answered by prakash jain last updated on 02/Oct/16

$$\mathrm{answer}\mathrm{in}\mathrm{comments}$$

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