If $ \frac{{ }^{\mathrm{n}} \mathrm{C}_{0}}{2}-\frac{{ }^{\mathrm{n}} \mathrm{C}_{1}}{3}+\frac{{ }^{\mathrm{n}} \mathrm{C}_{2}}{4}+\ldots . .+(-1)^{\mathrm{n}} \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}}{\mathrm{n}+2}=\frac{1}{2001 \times 2000} $, then $ \frac{(n+1)}{1000} $ is..

Question

If $ \frac{{ }^{\mathrm{n}} \mathrm{C}_{0}}{2}-\frac{{ }^{\mathrm{n}} \mathrm{C}_{1}}{3}+\frac{{ }^{\mathrm{n}} \mathrm{C}_{2}}{4}+\ldots . .+(-1)^{\mathrm{n}} \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}}{\mathrm{n}+2}=\frac{1}{2001 \times 2000} $, then $ \frac{(n+1)}{1000} $ is..,

MARKS App · Accepted Answer

( 1 - x ) n = C 0 n - C 1 n x + C 2 n x 2 - . . . . . . ( - 1 ) n C n n x n multiplying both sides by x x 1 - x n = C 0 n x - C 1 n x 2 + C 2 n x 3 - . . . . . . . - 1 n C n n x n + 1 Integrating both sides ∫ 0 1 x 1 - x n = ∫ 0 1 n C 0 x - n C 1 x 2 + n C 2 x 3 + ..... + n C n x n + 1 1 n + 1 n + 2 = n C 0 2 - n C 1 3 + n C 2 4 ..... ⇒ 1 n + 1 n + 2 = 1 2 0 0 0 × 2 0 0 1 i.e. n = 1 9 9 9

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