The value of ${ C }_{ 0 }^{ 2 }+3{ C }_{ 1 }^{ 2 }+5{ C }_{ 2 }^{ 2 }+.....$ to $n+1$ terms is

The general term of above series is Tr=(2r−1)C2r

Therefore, the last term of above series is Tn+1=(2(n+1)−1)C2n=(2n+1)C2n

C20+3C21+5C22+...+(2n+1)C2n

=∑r=nr=0(2r+1)C2r

=2∑r=nr=0rC2r+∑r=nr=0C2r

=2S1+S2

The binomial expansion of (1+x)n and (1+1x)n is given by

(1+x)n=C0+C1x+C2x2+....+Cnxn=∑r=nr=0Crxr ....[1]

(1+1x)n=C0+C11x+C21x2+....+Cn1xn=∑r=nr=0Cr1xr ....[2]

To find the value of S2, we can multiply above two equations and then compare the coefficients of terms independent of x.

Multiplying equation [1] and [2], we get

(1+x)n(1+1x)n=∑r=nr=0Crxr.∑r=nr=0Cr1xr

⟹(1+x)2nxn=∑r=nr=0Crxr.∑r=nr=0Cr1xr

⟹(1+x)2nxn=C20+C21+C22+...+C2n + terms containing x

Therefore, comparing coefficients of x0 in L.H.S and R.H.S, we get

Coefficients of x0 in R.H.S = Coefficients of x0 in L.H.S

⟹C20+C21+C22+...+C2n= coefficient of x0 in (1+x)2nxn

⟹C20+C21+C22+...+C2n=2nCn=S2 ....[3]

(1+1x)n=C0+C11x+C21x2+....+Cn1xn

Differentiating with respect to x, we get

⟹n(1+1x)n−1(−1x2)=0+C1−1x2+C2−2x3+....+Cn−nxn+1

⟹n(1+1x)n−1(1x)=C11x+C22x2+....+Cnnxn ....[4]

To find the value of S1, we can multiply equations [1] and [4], and then compare the coefficients of terms independent of x.

Multiplying equation [1] and [4], we get

n(1+1x)n−1(1x)(1+x)n=(C11x+C22x2+....+Cnnxn)(C0+C1x+C2x2+....+Cnxn)

⟹n(x+1)2n−1xn=(C11x+C22x2+....+Cnnxn)(C0+C1x+C2x2+....+Cnxn)

Therefore, comparing coefficients of x0 in L.H.S and R.H.S, we get

Coefficients of x0 in R.H.S = Coefficients of x0 in L.H.S

⟹C21+2C22+3C23+...+nC2n= coefficient of x0 in n(x+1)2n−1xn

⟹C21+2C22+3C23+...+nC2n=n(2n−1Cn)=S1

Therefore,

2S1+S2=2n(2n−1Cn)+2nCn

=2n(2n−1Cn)+2(2n−1Cn)

=2(n+1)(2n−1Cn)

Therefore, the answer is option C.