Problem Solving

hi... can anyone help me to find the sum of this series?

a + a^2 + a^4 + a^8 +... + a^(2^(n-1))= Sn

so what is the formula for Sn?

samuel minato wrote:hi... can anyone help me to find the sum of this series?

a + a^2 + a^4 + a^8 +... + a^(2^(n-1))= Sn

so what is the formula for Sn?

consider a^2 as x

so Sn = a + x + x^2 + .. + x^(n-1)

or Sn = a + GP(x)

Sum of a GP = b(r^n - 1)/(r-1) if r > 1 and b(1 - r^n)/(1 - r) if r < 1. Here r is the common ratio and b is the initial term

or Sn = a + b(r^n - 1)/(r - 1) if r > 1 and a + b(1 - r^n)/(1 - r) if r < 1

In our case b = r = a^2, n in our case is (n - 1)

so Sn = a + a^2(a^2(n-1) - 1)/(a^2 - 1) if r > 1 and a + a^2(1 - a^2(n-1))/(1 - a^2) if r < 1

thank you...