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.
isnt this an arithmetic series of value r=1/5 ? Isnt there a formula for it?
I think I got it. I'll have my answer shortly
Ok I have all my work down in the space but I have a spot that says:
Answer_________ What do they want me to put in that spot. I know "the answer" duh but I have all the steps and my work above that and I don't know what the "answer" is I don't want to do ALL that work and leave the answer place blank. Please tell me that made sense.
To 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, we will use mathematical induction.
Step 1: Base Case
Let's first check the base case when n = 1:
(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 equation holds true for the base case.
Step 2: Inductive Hypothesis
Assume that the statement is true for a positive integer k:
(1/5) + (1/5^2) + (1/5^3) + ... + (1/5^k) = (1/4)(1 - 1/5^k)
Step 3: Inductive Step
We need to show that if the statement is true for k, then it must also be true for k+1.
Now, consider the left-hand side of the equation when n = k + 1:
(1/5) + (1/5^2) + (1/5^3) + ... + (1/5^k) + (1/5^(k+1))
By the inductive hypothesis, we can replace the sum up to the kth term with (1/4)(1 - 1/5^k):
(1/4)(1 - 1/5^k) + (1/5^(k+1))
Let's simplify this expression:
= (1/4)(1 - 1/5^k) + (1/5^(k+1))
= (1/4)(1 - 1/5^k) + (1/5)(1/5^k) [Since 1/5^(k+1) = (1/5)(1/5^k)]
= (1/4)(1 - 1/5^k) + (1/4)(1/5^k) [Simplifying (1/5)*(1/5^k) = (1/4)(1/5^k)]
Now, apply the common denominator 4 to the two terms being added:
= (1/4)(4/4)(1 - 1/5^k) + (1/4)(1/5^k)
= (1/4)[4(1 - 1/5^k) + 1/5^k]
= (1/4)[4 - 4/5^k + 1/5^k]
= (1/4)[4 + (-4/5^k + 1/5^k)]
Now, simplify the expression inside the square brackets:
= (1/4)[4 + (-3/5^k)]
= (1/4)(4 - 3/5^k)
Notice that (4 - 3/5^k) is equal to 1 - 1/5^(k+1):
= (1/4)(1 - 1/5^(k+1))
This is the same as the right-hand side of the equation we want to prove.
Therefore, if the statement holds true for k, it also holds true for k+1.
Step 4: Conclusion
By using mathematical induction, we have shown 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.