If (x)<1, and rth term of a series is 1 + x+x2+…+xr−1, then sum to n terms of the series is
n+(n+1)x−xn+1(1−x)2
n−(n+1)x+xn+1(1−x)2
(n+1)x−xn+1−n(1−x)2
n−(n+1)x−xn+1(1−x)2
tr=1+x+x2+…xr−1=1−xr1−x
Now, ∑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