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Question

Mathematics Question on Binomial theorem

If the value of C0+2C1+3C2++(n+1)Cn=576C_0 + 2 \cdot C_1 + 3 \cdot C_2 + \dots+ (n+1) \cdot C_n=576 ,then n is equal to

A

7

B

5

C

6

D

9

Answer

7

Explanation

Solution

Given, C0+2C1+3C2++(n+1)Cn=576C_{0}+2 C_{1}+3 C_{2}+\ldots+(n+1) C_{n}=576
We know that,
(1+x)n=nC0+nC1x+nC2x2++nCnxn(1+x)^{n}={ }^{n} C_{0}+{ }^{n} C_{1} x+{ }^{n} C_{2} x^{2}+\ldots+{ }^{n} C_{n} x^{n}
x(1+x)n=nC0x+nC1x2+nC2x3++nCnxn+1\Rightarrow x(1+x)^{n}={ }^{n} C_{0} x+{ }^{n} C_{1} x^{2}+{ }^{n} C_{2} x^{3}+\ldots +{ }^{n} C_{n} x^{n+1}
On differentiating w.r.t. xx, we get
(1+x)n+xn(1+x)n1(1+x)^{n}+x \cdot n(1+x)^{n-1}
=nC0+2nC1x+3nC2x2++(n+1)nCnxn={ }^{n} C_{0}+2 \cdot{ }^{n} C_{1} \cdot x+3{ }^{n} C_{2} x^{2}+\ldots+(n+1){ }^{n} C_{n} x^{n}
On putting n=1n=1, we get
2n+n2n1=nC0+2nC1+3nC1++(n+1)nCn2^{n}+n \cdot 2^{n-1}={ }^{n} C_{0}+2 \cdot{ }^{n} C_{1}+3 \cdot{ }^{n} C_{1}+\ldots +(n+1){ }^{n} C_{n}
2n1(n+2)=576\Rightarrow 2^{n-1}(n+2)=576 \,\,\,\,\,\, (given)
2n1(n+2)=26×9=2(71)(7+2)\Rightarrow 2^{n-1}(n+2)=2^{6} \times 9=2^{(7-1)} \cdot(7+2)
On comparing, we get n=7n=7