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Question

Question: \[2^{10} - 1\]...

210−12^{10} - 1

A

101

B

C0Cr+C1Cr+1+C2Cr+2+....+Cn−rCnC_{0}C_{r} + C_{1}C_{r + 1} + C_{2}C_{r + 2} + .... + C_{n - r}C_{n}

C

(2n)!(n−r)!(n+r)!\frac{(2n)!}{(n - r)!(n + r)!}

D

n!(−r)!(n+r)!\frac{n!}{( - r)!(n + r)!}

Answer

(2n)!(n−r)!(n+r)!\frac{(2n)!}{(n - r)!(n + r)!}

Explanation

Solution

c)

Sol. =2nCn=(2n)!(n!)2=^{2n}C_{n} = \frac{(2n)!}{(n!)^{2}}

(1−9x+20x2)−1=[(1−5x)(1−4x)]−1(1 - 9x + 20x^{2})^{- 1} = \lbrack(1 - 5x)(1 - 4x)\rbrack^{- 1}

∴ =1(1−5x)(1−4x)=51−5x−41−4x= \frac{1}{(1 - 5x)(1 - 4x)} = \frac{5}{1 - 5x} - \frac{4}{1 - 4x}

=5(1−5x)−1−4(1−4x)−1= 5(1 - 5x)^{- 1} - 4(1 - 4x)^{- 1}

=5[1+5x+(5x)2+...+(5x)n+...]= 5\lbrack 1 + 5x + (5x)^{2} + ... + (5x)^{n} + ...\rbrack

=−4[1+4x+(4x)2.....+(4x)n+...]- 4\lbrack 1 + 4x + (4x)^{2}..... + (4x)^{n} + ...\rbrack.