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Using mathematical induction, prove that the sum of the first ๐’ positive integers isequal to ๐’(๐’+๐Ÿ)๐Ÿ .

Question

Using mathematical induction, prove that the sum of the first ๐’ positive integers isequal to ๐’(๐’+๐Ÿ)๐Ÿ .

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Solution

Sure, here is the proof using mathematical induction:

Step 1: Base Case First, we need to prove the statement is true for n=1. The sum of the first 1 positive integer is 1, and 1*(1+1)/2 = 1. So, the statement is true for n=1.

Step 2: Inductive Step Next, we assume the statement is true for some arbitrary positive integer k. That is, we assume that the sum of the first k positive integers is k*(k+1)/2.

Step 3: Inductive Hypothesis We need to prove that the statement is true for n=k+1. The sum of the first k+1 positive integers is (k+1) + k*(k+1)/2 = (k+1)*(k+2)/2.

Step 4: Conclusion Since the statement is true for n=1 and if it's true for n=k then it's true for n=k+1, by the principle of mathematical induction, the statement is true for all positive integers n.

This problem has been solved

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A student is trying to prove a statement ๐‘„(๐‘›)๐‘„(๐‘›) for all integers ๐‘›โ‰ฅ2๐‘›โ‰ฅ2 by strong induction. As an induction step they have proved that, for all integers ๐‘˜โ‰ฅ3๐‘˜โ‰ฅ3, if ๐‘„(๐‘˜โˆ’1)๐‘„(๐‘˜โˆ’1) then ๐‘„(๐‘˜+1)๐‘„(๐‘˜+1). What is the minimum number of base steps they will need to finish their proof?

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