show that the coefficient of the middle term in the expansion of (1+x)^2n is equal to the sum of the coefficients of two middle terms in the expansion (1+x)^2n-1.

Question

show that the coefficient of the middle term in the expansion of
(1+x)^2n is equal to the sum of the coefficients of two middle
terms in the expansion (1+x)^2n-1

Anuradha Sharma · Accepted Answer

The index 2n in 1+x2n is even
Middle term = T2n+22=Tn+1=2nCn1nxn=2nCnxnNow coefficient of middle term in 1+x2n=2nCn
The index  2n-1 in  1+x2n-1 is odd
So middle term  are T2n-1+12 and the next term i
e
, Tn and Tn+1 Tn=Tn-1+1=2n-1Cn-1×1n×xn-1=2n-1Cn-1
xn-1and Tn+1=2n-1Cn×1n-1×xn=2n-1Cn
xnNow sum of coefficient of middle terms in 1+x2n-1=2n-1Cn-1+2n-1Cn=2n-1+1Cn=2nCn=coefficient of middle term in 1+x2nUsing property nCr+nCn-1=n+1Cr