The value of - r -0 n -1 r n C r - r -2 is equal toA- 1-n-1 nB- 1-n-2 nC- 1-n-1n-2D- 1-nn-1n-2

The correct option is C C_0x C_1x^2 + C_2x^3 ....... = x(1 x)^n nbsp; Integrating both sides , within the limits 0 to 1 nbsp; [ ( C_0 x^2/2 nbsp;C_1 ...

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Standard XII

Mathematics

Method of Difference

The value of ...

Question

The value of $n \sum r = 0 (- 1)^{r} \frac{^{n} C_{r}}{r + 2}$ is equal to .

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Solution

The correct option is C

$C_{0} x - C_{1} x^{2} + C_{2} x^{3} . . . . . . . = x (1 - x)^{n}$

Integrating both sides , within the limits 0 to 1

${[(\frac{C_{0} x^{2}}{2} - \frac{C_{1} x^{3}}{3} + \frac{C_{2} x^{4}}{4} - . . . . . . .)]}_{0}^{1} = \int_{0}^{1} x (1 - x)^{n} d x$

$\int_{0}^{1} (1 - x) x^{n} d x = {[\frac{x^{n - 1}}{n + 1} - \frac{x^{n + 2}}{n + 2}]}_{0}^{1}$

$\Rightarrow$ $\frac{C_{0}}{2} - \frac{C_{1}}{3} + \frac{C_{2}}{4} - . . . . = \frac{1}{(n + 1)} - \frac{1}{(n + 2)} = \frac{1}{(n + 1) (n + 2)} .$

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The value of n∑r=0(−1)rnCrr+2 is equal to .

The correct option is C C0x−C1x2+C2x3.......=x(1−x)n Integrating both sides , within the limits 0 to 1 [(C0x22−C1x33+C2x44−.......)]10=∫10x(1−x)ndx ∫10(1−x)xndx=[xn−1n+1−xn+2n+2]10 ⇒ C02−C13+C24−....=1(n+1)−1(n+2)=1(n+1)(n+2).

The value of $n \sum r = 0 (- 1)^{r} \frac{^{n} C_{r}}{r + 2}$ is equal to .