(1+x)^n=nc0 + nc1(x) + nc2(x^2)+.................................+ncn(x^n)

we know sum of the series is

nc0+nc1(2) +nc2(2^2)+nc3(2^3)+.......................+ncn(2^n)

we can find da sum by plaxing 2 in place of x in the expansion (1+x)^n

i.e 3^n=6561

3^8 =6561

hence n=8

middle term is greatest i.e

ncn/2(2^n/2)

n=8

ans is 1120

reply if i am right