prove that the statement: 1/5+1/(5^2)+1/(5^3)+...+1/(5^n)=1/4(1-1/5n) is true for all positive integers n.?

Question

step by step

Answer 1

1/5 + 1/(5^2) + ... + (1/5^n) = 1/4*(1-1/5^n)

By induction:

n = 1:
1/5 = 1/4(1-1/5)
1/5 = 1/4 * 4/5 = 1/5 => TRUE for n = 1;

Suppose that this is true for n = k, we shall show that it is also true for n = k+1, that means:
1/5 + 1/(5^2) + ... + 1/5^(k+1) = 1/4*[1-1/5^(k+1)]

For n = k
1/5 + 1/(5^2) + ... + (1/5^k) = 1/4*(1-1/5^k)
n = k+1
1/5 + 1/(5^2) + ... + (1/5^k) + 1/5^(k+1)
= 1/4 (1-1/5^k) + 1/5^(k+1)
= 1/4 (5^k - 1)/5^k + 1/5^(k+1)
= 1/5^k*[(5^k - 1)/4 + 1/5]
= 1/5^k *[5^(k+1)-5 + 4]/[4*5]
= 1/4*1/5^(k+1)*[5^(k+1) - 1]
= 1/4 * [1 - 1/5^(k+1)]
=> Done!

Answer 2

1/5 + 1/(5^2) + ... + 1/(5^n) = 1/4(1 - 1/5^n)

If n=1

1/4(1 - 1/5) = 1/4 * 4/5 = 1/5

Let's assume for n and prove for n+1

1/5 + 1/5^2 + ... + 1/5^n + 1/5^(n + 1) = 1/4(1 - 1/5^n) +
+ 1/5^(n + 1) = 1/4 [(5^n - 1)/5^n] + 1/5^(n+1) =
= 5/4[(5^n - 1)/5^(n + 1)] + 1/5^(n + 1) =
= 1/5^(n + 1) * [5/4(5^n - 1) + 1] = 1/5^(n + 1)* [5^(n + 1) - 5 +
+ 4]/4 = 1/4 {1/5^(n + 1) * [5^(n + 1) - 1]} =
= 1/4 * [1 - 1/5^(n + 1)]

Answer 3

what is the "statement?" That's just a series.