carameled
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Prove by math induction that
n
sigma 3i + 1 = n/2 (3n + 5)
i = n
n
sigma 3i + 1 = n/2 (3n + 5)
i = n
The discussion centers on proving the mathematical statement using induction, specifically the formula $$\sum_{i=1}^{n}(3i+1)=\frac{n}{2}(3n+5)$$. Participants confirm the base case \(P_1\) by evaluating $$\sum_{i=1}^{1}(3i+1)=\frac{1}{2}(3(1)+5)$$, establishing its truth. The importance of understanding the induction hypothesis and the base case is emphasized as fundamental concepts in mathematical induction. Clarifications regarding the correct indexing in the summation are also addressed.
PREREQUISITESStudents of mathematics, educators teaching algebra and calculus, and anyone interested in mastering mathematical proofs and induction techniques.
MarkFL said:I think what you mean is the induction hypothesis \(P_n\):
$$\sum_{i=1}^{n}\left(3i+1\right)=\frac{n}{2}(3n+5)$$
The first thing we want to do is confirm the base case \(P_1\) is true:
$$\sum_{i=1}^{1}\left(3i+1\right)=\frac{1}{2}(3(1)+5)$$
Is this true?
carameled said:Wow, well I'm just asking for the prove with math induction. I don't understand any of that..
MarkFL said:You don't understand what an induction hypothesis is, or demonstrating the truth of the base case? These are fundamental to induction. What method have you been taught?