Question

Please solve this, need help with this

Answer 1

Employing mathematical induction establishes that the given geometric series is equal to
`\(1 - (1)/(5^n)\)` for all positive integers (n), affirming the validity of the statement.

To prove the given statement by induction, we need to follow these steps:

1. Base Case: Show that the statement is true for n = 1.

2. Inductive Step: Assume that the statement is true for some arbitrary positive integer k (inductive hypothesis), and then prove that it is also true for k + 1.

Let's proceed with the proof:

Base Case (n = 1):

For n = 1, the left side of the equation is:

4/5

The right side of the equation is:

`\[ 1 - (1)/(5^1) = (5)/(5) - (1)/(5) = (4)/(5) \]`

The left side equals the right side for n = 1, so the base case holds.

Inductive Step:

Assume that the statement is true for some arbitrary positive integer k:

`\[ 4\left((1)/(5) + (1)/(5^2) + (1)/(5^3) + \ldots + (1)/(5^k)\right) = 1 - (1)/(5^k) \]`

Now, we need to prove that it holds for k + 1:

`\[ 4\left((1)/(5) + (1)/(5^2) + (1)/(5^3) + \ldots + (1)/(5^k) + (1)/(5^(k+1))\right) = 1 - (1)/(5^(k+1)) \]`

Let's add
`\((4)/(5^(k+1))\)` to both sides of the inductive hypothesis:

`\[ 4\left((1)/(5) + (1)/(5^2) + (1)/(5^3) + \ldots + (1)/(5^k) + (1)/(5^(k+1))\right) + (4)/(5^(k+1)) = 1 - (1)/(5^k) + (4)/(5^(k+1)) \]`

Combine the terms on the left side:

`\[ 4\left((1)/(5) + (1)/(5^2) + (1)/(5^3) + \ldots + (1)/(5^k) + (1)/(5^(k+1))\right) + (4)/(5^(k+1)) = (4)/(5) + (4)/(5^2) + (4)/(5^3) + \ldots + (4)/(5^k) + (4)/(5^(k+1)) \]`

Now, factor out the common factor of 4/5:

`\[ (4)/(5) \left(1 + (1)/(5) + (1)/(5^2) + \ldots + (1)/(5^k) + (1)/(5^(k+1))\right) + (4)/(5^(k+1)) = (4)/(5) \left((5)/(5) - (1)/(5^k)\right) + (4)/(5^(k+1)) \]`

Simplify the expression:

`\[ (4)/(5) \left(1 - (1)/(5^k)\right) + (4)/(5^(k+1)) = (4)/(5) - (4)/(5^(k+1)) + (4)/(5^(k+1)) \]`

Combine the terms on the right side:

`\[ (4)/(5) - (4)/(5^(k+1)) + (4)/(5^(k+1)) = (4)/(5) \]`

Therefore, the statement holds for k + 1.

By induction, the statement is proven to be true for all positive integers n.