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Question: The value of (n+2)C<sub>0­</sub>2<sup>n+1</sup> – (n+1)C<sub>1­</sub>2<sup>n</sup>+ nC<sub>2­</sub>2...

The value of (n+2)C0­2n+1 – (n+1)C1­2n+ nC2­2n-1+ . . . is equal to

A

4n

B

4

C

2n+4

D

4(1+n)

Answer

4(1+n)

Explanation

Solution

Since (x-1)n = C0­xn – C1­xn-1+ . . .

x2(x-1)n = C0­xn+2 – C1­xn+1+ . . .

Differentiating w.r.t. x we get

2x(x-1)n - x2 n(x-1)n -1= (n+2)C0­xn+1 – (n+1) C1­xn+ . . .

Put x=2 and the result is 4(n+1)