The-value-of-the-sum-of-the-series-3-n-C-0-8-n-C-1-13-n-C-2-18-n-C-3-1-n-3-5n-n-C-n-

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

The value of the sum of the series

3^{n} C_{0} - 8^{n} C_{1} + 13^{n} C_{2} - 18^{n} C_{3} + \dots + {(- 1)}^{n} (3 + 5 n)^{n} C_{n}]

Commented by Yozzias last updated on 02/Oct/16

\underset{0}{\sum^{n}} (\begin{matrix} n \\ r \end{matrix}) {(- 1)}^{r} (3 + 5 r) = 3 \underset{0}{\sum^{n}} (\begin{matrix} n \\ r \end{matrix}) {(- 1)}^{r} + 5 \underset{0}{\sum^{n}} (\begin{matrix} n \\ r \end{matrix}) {(- 1)}^{r} r

Let {(1 - x)}^{n} = \underset{0}{\sum^{n}} (\begin{matrix} n \\ r \end{matrix}) {(- 1)}^{r} x^{r} \dots . . (1)

Differentiating \Rightarrow - n {(1 - x)}^{n - 1} = \underset{0}{\sum^{n}} (\begin{matrix} n \\ r \end{matrix}) r {(- 1)}^{r} x^{r - 1}

x = 1 \Rightarrow 0 = \underset{0}{\sum^{n}} (\begin{matrix} n \\ r \end{matrix}) {(- 1)}^{r} r \dots \dots . (2)

Also, x = 1 in (1) \Rightarrow \underset{0}{\sum^{n}} (\begin{matrix} n \\ r \end{matrix}) {(- 1)}^{r} = 0 .

∴ \underset{0}{\sum^{n}} (\begin{matrix} n \\ r \end{matrix}) {(- 1)}^{r} (3 + 5 r) = 3 \times 0 + 0

\underset{0}{\sum^{n}} (\begin{matrix} n \\ r \end{matrix}) {(- 1)}^{r} (3 + 5 r) = 0

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

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