1 Answer

Leastprivilege · Answer 1 · 2024-06-26T18:08:04+0000

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:

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