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Question: If $(1+x)^n=C_0+C_1x+C_2x^2+C_3x^3+.......+C_nx^n$, then find the sum $C_0C_r+C_1C_{r+1}+C_2C_{r+2}+...

If (1+x)n=C0+C1x+C2x2+C3x3+.......+Cnxn(1+x)^n=C_0+C_1x+C_2x^2+C_3x^3+.......+C_nx^n, then find the sum C0Cr+C1Cr+1+C2Cr+2+.......+Cn−rCnC_0C_r+C_1C_{r+1}+C_2C_{r+2}+.......+C_{n-r}C_n

A

(2nn−r)\binom{2n}{n-r}

B

(2nn+r)\binom{2n}{n+r}

C

(nr)\binom{n}{r}

D

(nn−r)\binom{n}{n-r}

Answer

The sum is (2nn−r)\binom{2n}{n-r} or equivalently (2nn+r)\binom{2n}{n+r}.

Explanation

Solution

The given sum is S=C0Cr+C1Cr+1+C2Cr+2+.......+Cn−rCnS = C_0C_r+C_1C_{r+1}+C_2C_{r+2}+.......+C_{n-r}C_n. This can be written in summation notation as: S=∑k=0n−rCkCk+rS = \sum_{k=0}^{n-r} C_k C_{k+r}

Since CkC_k represents the binomial coefficients, Ck=(nk)C_k = \binom{n}{k}. Substituting this into the sum: S=∑k=0n−r(nk)(nk+r)S = \sum_{k=0}^{n-r} \binom{n}{k} \binom{n}{k+r}

Using the property of binomial coefficients (nm)=(nn−m)\binom{n}{m} = \binom{n}{n-m}, we can rewrite (nk+r)\binom{n}{k+r} as (nn−(k+r))=(nn−k−r)\binom{n}{n-(k+r)} = \binom{n}{n-k-r}. So, the sum becomes: S=∑k=0n−r(nk)(nn−k−r)S = \sum_{k=0}^{n-r} \binom{n}{k} \binom{n}{n-k-r}

This sum is the coefficient of xn−rx^{n-r} in the expansion of (1+x)n×(1+x)n=(1+x)2n(1+x)^n \times (1+x)^n = (1+x)^{2n}. The expansion of (1+x)2n(1+x)^{2n} is given by: (1+x)2n=∑j=02n(2nj)xj(1+x)^{2n} = \sum_{j=0}^{2n} \binom{2n}{j} x^j

The coefficient of xn−rx^{n-r} in (1+x)2n(1+x)^{2n} is (2nn−r)\binom{2n}{n-r}. Therefore, the sum S=(2nn−r)S = \binom{2n}{n-r}.

Using the identity (NK)=(NN−K)\binom{N}{K} = \binom{N}{N-K}, we can also write: (2nn−r)=(2n2n−(n−r))=(2nn+r)\binom{2n}{n-r} = \binom{2n}{2n-(n-r)} = \binom{2n}{n+r}. Thus, the sum can also be expressed as (2nn+r)\binom{2n}{n+r}.