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Question: If C<sub>0</sub>, C<sub>1</sub>, C<sub>2</sub>,.....,C<sub>n</sub> are their usual meaning, then\(\f...

If C0, C1, C2,.....,Cn are their usual meaning, thenC0n(n+1)\frac{C_{0}}{n(n + 1)}–C1(n+1)(n+2)\frac{C_{1}}{(n + 1)(n + 2)}+C2(n+2)(n+3)\frac{C_{2}}{(n + 2)(n + 3)}+ .... to (n+1) terms is equal to–

A

∫01xn+1(1−x)n+1dx\int_{0}^{1}{x^{n + 1}(1 - x)^{n + 1}dx}

B

∫01xn(1−x)n+1dx\int_{0}^{1}{x^{n}(1 - x)^{n + 1}dx}

C

∫01xn−1(1−x)n+1dx\int_{0}^{1}{x^{n - 1}(1 - x)^{n + 1}dx}

D

None of these

Answer

∫01xn−1(1−x)n+1dx\int_{0}^{1}{x^{n - 1}(1 - x)^{n + 1}dx}

Explanation

Solution

(1–x)n = C0 – C1x + C2x2 – C3x3 + .......+ (–1)n Cnxn

Multiplying by xn–1 to both side

xn–1 (1 –x)n = C0 xn–1 –C1xn + C2xn+1 .....

...+ (–1)n Cnx2n–1

Now multiplying with (1 – x) to both sides

xn–1 (1 – x)n+1 = (C0 xn–1 – C1 xn + C2xn +1.......

+ (–1)n Cnx2n–1) (1 – x)

Now integrating w.r.t. x from 0 to 1

∫01xn−1(1−x)n+1dx\int_{0}^{1}{x^{n - 1}(1 - x)^{n + 1}dx}= C0n(n+1)\frac{C_{0}}{n(n + 1)}– C1(n+1)(n+2)\frac{C_{1}}{(n + 1)(n + 2)}+

C2(n+2)(n+3)\frac{C_{2}}{(n + 2)(n + 3)}.........