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Mathematics Question on binomial expansion formula

The coefficient of xnx^n in the polynomial (x+nC0)(x+3.nC1)(x+5.nC2)....(x+(2n+1)nCn)\left(x+^{n}C_{0}\right)\left(x+3. ^{n}C_{1}\right)\left(x+5. ^{n}C_{2}\right) .... \left(x+\left(2n+1\right)^{n}C_{n}\right) is

A

n.2nn . 2^n

B

n.2n+1n . 2^{n + 1}

C

(n+1).2n(n + 1) . 2^n

D

(n+1).2n+1(n + 1) . 2^{n+1}

Answer

(n+1).2n(n + 1) . 2^n

Explanation

Solution

(x+nC0)(x+3.nC1)(x+5.nC2)....(x+(2n+1).nCn)\left(x+^{n}C_{0}\right)\left(x+3. ^{n}C_{1}\right)\left(x+5. ^{n}C_{2}\right) .... \left(x+\left(2n+1\right).^{n}C_{n}\right) = x^{n+1}+x^{n}\left\\{^{n}C_{0}+3. ^{n}C_{1}+5. ^{n}C_{2}+..... +\left(2n+1\right).^{n}C_{n}\right\\}+..... Coeff. of xn=nC0+3.nC1+5.nC2+.....+(2n+1).nCnx^{n} = ^{n}C_{0} +3. ^{n}C_{1}+5. ^{n}C_{2}+..... +\left(2n+1\right).^{n}C_{n} =1+(nC1+2.nC1)+(nC2+4.nC2)+....+(nCn+2n.nCn) = 1+ \left(^{n}C_{1} +2. ^{n}C_{1}\right)+\left(^{n}C_{2}+4. ^{n}C_{2}\right) + ....+\left(^{n}C_{n}+2n. ^{n}C_{n}\right) =(1+nC1+.....+nCn)+2(nC1+2nC2+....+n.nCn)= \left(1+^{n}C_{1}+..... + ^{n}C_{n}\right)+2\left(^{n}C_{1} + 2^{n}C_{2} + .... +n. ^{n}C_{n}\right) =2n+2[n+2.n(n−1)2!+3.n(n−1)(n−2)3!+...+n.1]= 2^{n} +2\left[n+2. \frac{n\left(n-1\right)}{2!}+3. \frac{n\left(n-1\right)\left(n-2\right)}{3!}+...+ n.1\right] =2n+2n[1+n−1C1+n−1C2+.....+n−1Cn−1]= 2^{n} +2n \left[1+^{n-1}C_{1}+^{n-1}C_{2}+ ..... +^{n-1}C_{n-1}\right] =2n+2n.2n−1=2n(1+n)=(n+1).2n= 2^{n}+2^{n}. 2^{n-1} = 2^{n} \left(1+n\right) = \left(n+1\right) . 2^{n}