The value of $\frac{C_{0}}{1.3} - \frac{C_{1}}{2.3} + \frac{C_{2}}{3.3} - \frac{C_{3}}{4.3} + . . . + {(- 1)}^{n} \frac{C_{n}}{(n + 1) . 3}$ is

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$\frac{3}{n + 1}$

$\frac{n + 1}{3}$

$\frac{1}{3 (n + 1)}$

None of these

answer is C.

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Detailed Solution

We have to find the value of

$\frac{C_{0}}{1.3} - \frac{C_{1}}{2.3} + \frac{C_{2}}{3.3} - \frac{C_{3}}{4.3} + \dots + (- 1)^{n} \frac{C_{n}}{(n + 1) \cdot 3}$

we have,

$(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots . . + C_{n} x^{n}$

Integrating both sides w.r.t.x , from -1 to 0, we get

$\begin{array}{l} \int_{- 1}^{0} (1 + x)^{n} d x = \int_{- 1}^{0} (C_{0} + C_{1} x + C_{2} x^{2} + \dots . + C_{n} x^{n}) d x \\ \frac{(1 + x)^{n + 1}}{n + 1} \int_{- 1}^{0} = {(\cos x + \frac{C_{1} x^{2}}{2} + \dots . . + \frac{C_{n x^{n} + 1}}{n + 1})}_{- 1}^{0} \\ \Rightarrow C_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{2} + \dots . + (- 1)^{n} \frac{C_{n}}{n + 1} = \frac{1}{n + 1} \\ ∴ The given expression = \frac{1}{3} (C_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{2} \dots \dots .) \\ = \frac{1}{3} \cdot \frac{1}{n + 1} = \frac{1}{3 (n + 1)} \end{array}$