The problem is:Prove that the statement 1/5+1/5^2+1/5^3+...1/5^n=1/4(1-1/5^n) is true for all positive integers n. Write your proof in the space below.

How do I start this? I have looked at the only example in the book but it did not help me.

Any help in this would be great!!

To start proving this statement, you can use mathematical induction.

1. Base case (n = 1):
Substitute n = 1 into the statement:
1/5 = 1/4(1 - 1/5^1)
1/5 = 1/4(1 - 1/5)
1/5 = 1/4(4/5)
1/5 = 1/5
The base case holds true.

2. Inductive step:
Assume that the statement is true for some positive integer k, i.e., assume 1/5 + 1/5^2 + 1/5^3 + ... + 1/5^k = 1/4(1 - 1/5^k).

We need to prove that the statement is also true for k + 1, i.e., 1/5 + 1/5^2 + 1/5^3 + ... + 1/5^k + 1/5^(k + 1) = 1/4(1 - 1/5^(k + 1)).

Start with the left-hand side (LHS) of the k + 1 expression:
LHS = 1/5 + 1/5^2 + 1/5^3 + ... + 1/5^k + 1/5^(k + 1)

Combine the terms to create a common denominator:
LHS = (1/5^k) * (1/5^0 + 1/5^1 + 1/5^2 + ... + 1/5^(k - 1)) + 1/5^(k + 1)
LHS = (1/5^k) * (1 - 1/5^k) + 1/5^(k + 1)

Now, let's simplify the right-hand side (RHS) of the k + 1 expression:
RHS = 1/4(1 - 1/5^(k + 1))

Expand the RHS:
RHS = 1/4 - 1/(4 * 5^(k + 1))

Now, let's focus on the LHS again and simplify further:
LHS = (1/5^k) * (1 - 1/5^k) + 1/5^(k + 1)
= (1/5^k) - (1/5^(2k)) + 1/5^(k + 1)

Now, to confirm that the LHS is equal to the RHS, we need to show that:
(1/5^k) - (1/5^(2k)) + 1/5^(k + 1) = 1/4 - 1/(4 * 5^(k + 1))

Let's manipulate the LHS to match the RHS:
(1/5^k) - (1/5^(2k)) + 1/5^(k + 1) = (5 * 5^k - 5^(2k) + 5^k)/(5^(k + 1))

Simplify the numerator:
(5 * 5^k - 5^(2k) + 5^k) = (5^k) * (5 - 5^k + 1)

Simplify the numerator further:
(5^k) * (5 - 5^k + 1) = (5^k) * (6 - 5^k)

Plug the simplified numerator back into the expression:
(5^k * (6 - 5^k))/(5^(k + 1)) = (6 - 5^k)/5

We can observe that (6 - 5^k)/5 = 1 - 1/(5^(k + 1)).
This can be verified by multiplying both sides by 5 and simplifying.

Therefore, LHS = RHS.

Since the base case holds true and the inductive step has been proven, by mathematical induction, the statement 1/5 + 1/5^2 + 1/5^3 + ... + 1/5^n = 1/4(1 - 1/5^n) is true for all positive integers n.