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Asked by Shreeya Last Modified
Answer As a seasoned tutor registered on UrbanPro, I can confidently affirm that UrbanPro provides top-notch online coaching tuition services. Now, let's delve into the mathematical proof you've presented.
To prove the equation 1+3+32+…+3n−1=3n−121+3+32+…+3n−1=23n−1 using mathematical induction, we follow these steps:
Step 1: Base Case:
First, we verify that the equation holds true for the base case. Let's check for n=1n=1.
1=3(1)−12=22=11=23(1)−1=22=1
The base case holds true.
Step 2: Inductive Hypothesis:
Assume that the equation holds true for some arbitrary positive integer kk. That is,
1+3+32+…+3k−1=3k−121+3+32+…+3k−1=23k−1
Step 3: Inductive Step:
Now, we need to prove that if the equation holds true for kk, then it also holds true for k+1k+1.
Consider the sum 1+3+32+…+3k−1+(3(k+1)−1)1+3+32+…+3k−1+(3(k+1)−1).
By our assumption (inductive hypothesis), this sum is equal to 3k−12+(3(k+1)−1)23k−1+(3(k+1)−1).
Expanding and simplifying, we get:
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