If (x)<1, and rth term of a series is 1 + x+x2+…+xr−1, then sum to n terms of the series is

tr=1+x+x2+…xr−1=1−xr1−xNow, ∑r=1n tr=∑r=1n 1−xr1−x=n1−x−11−x∑r=1n xr=n1−x−11−xx1−xn1−x=1(1−x)2n−nx−x+xn+1=1(1−x)2n−(n+1)x+xn+1