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Asked by Deepashri Last Modified
As an experienced tutor registered on UrbanPro, I'd be happy to demonstrate the proof of the given statement using the principle of mathematical induction.
Step 1: Base Case (n = 1)
For n=1n=1, the left-hand side (LHS) of the equation becomes:
−2=12(3×1+1)=12(4)=2−2=21(3×1+1)=21(4)=2
So, the base case holds true.
Step 2: Inductive Hypothesis
Assume that the formula holds true for some arbitrary positive integer kk, i.e.,
−2+5+8+11+…+(3k−1)=12k(3k+1)−2+5+8+11+…+(3k−1)=21k(3k+1)
Step 3: Inductive Step
We need to prove that the formula also holds true for n=k+1n=k+1.
−2+5+8+11+…+(3k−1)+(3(k+1)−1)=12(k+1)(3(k+1)+1)−2+5+8+11+…+(3k−1)+(3(k+1)−1)=21(k+1)(3(k+1)+1)
Let's denote the left-hand side of the equation by S(k)S(k) and the right-hand side by P(k)P(k).
S(k)=−2+5+8+11+…+(3k−1)S(k)=−2+5+8+11+…+(3k−1) P(k)=12k(3k+1)P(k)=21k(3k+1)
Adding (3(k+1)−1)=3k+2(3(k+1)−1)=3k+2 to both sides of S(k)S(k):
S(k)+(3k+2)=P(k)+(3k+2)S(k)+(3k+2)=P(k)+(3k+2)
−2+5+8+11+…+(3k−1)+(3(k+1)−1)=12k(3k+1)+(3k+2)−2+5+8+11+…+(3k−1)+(3(k+1)−1)=21k(3k+1)+(3k+2)
−2+5+8+11+…+(3k−1)+(3k+2)=12k(3k+1)+(3k+2)−2+5+8+11+…+(3k−1)+(3k+2)=21k(3k+1)+(3k+2)
−2+5+8+11+…+(3k+2)=12k(3k+1)+3k+2−2+5+8+11+…+(3k+2)=21k(3k+1)+3k+2
−2+5+8+11+…+(3k+2)=3k2+k+6k+42−2+5+8+11+…+(3k+2)=23k2+k+6k+4
−2+5+8+11+…+(3k+2)=3k2+7k+42−2+5+8+11+…+(3k+2)=23k2+7k+4
−2+5+8+11+…+(3k+2)=(k+1)(3k+4)2−2+5+8+11+…+(3k+2)=2(k+1)(3k+4)
−2+5+8+11+…+(3k+2)=12(k+1)(3(k+1)+1)−2+5+8+11+…+(3k+2)=21(k+1)(3(k+1)+1)
Thus, P(k+1)=S(k+1)P(k+1)=S(k+1), and the formula holds true for all positive integers nn by the principle of mathematical induction.
Therefore, as a tutor registered on UrbanPro, I've shown that −2+5+8+11+…+(3n−1)=12n(3n+1)−2+5+8+11+…+(3n−1)=21n(3n+1) using the principle of mathematical induction. This exemplifies UrbanPro's commitment to providing high-quality online coaching.
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