# Math Help - geometric seies

1. ## geometric seies

Use a/k + a/k(1-1/k)+.....a/k(1-1/k)^n+....=a to show that
3/4 +3/16 + 3/64 + ....+ 3/4^n....= 1

2. Originally Posted by bigb
Use a/k + a/k(1-1/k)+.....a/k(1-1/k)^n+....=a to show that
3/4 +3/16 + 3/64 + ....+ 3/4^n....= 1
Given formula: $\frac{a}{k} \left[1 + \left(1 - \frac{1}{k}\right) + \, .... \, + \left(1 - \frac{1}{k}\right)^n + \, .... \right] = a$.

Required to find: $\frac{3}{4} \left(1 + \frac{1}{4} + \, .... \, + \frac{1}{4^{n-1}} + \frac{1}{4^n} + \, .... \right)$.

So substitute a = 1 and k = 4/3 into the given formula ......

3. Originally Posted by mr fantastic
Given formula: $\frac{a}{k} \left[1 + \left(1 - \frac{1}{k}\right) + \, .... \, + \left(1 - \frac{1}{k}\right)^n + \, .... \right] = a$.

Required to find: $\frac{3}{4} \left(1 + \frac{1}{4} + \, .... \, + \frac{1}{4^{n-1}} + \frac{1}{4^n} + \, .... \right)$.

So substitute a = 1 and k = 4/3 into the given formula ......
i get the series to be 15/16 and it is suppose to be 1

4. Originally Posted by bigb
i get the series to be 15/16 and it is suppose to be 1
Obviously the answer is 1 ..... because a = 1. I gave you that in my first reply.