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To solve the problem of proving the sum of the first (n) positive integers equals (\frac{n(n+1)}{2}) using mathematical induction, follow these steps:

Problem Statement

Prove that for any positive integer (n):
[1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}]

Proof by Mathematical Induction

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Step 1: Base Case ((n=1))

When (n=1):

• Left-hand side (LHS): (1)
• Right-hand side (RHS): (\frac{1(1+1)}{2} = \frac{2}{2} =1)

Since (LHS = RHS), the formula holds for (n=1).

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Step 2: Inductive Hypothesis

Assume the formula is true for some positive integer (k):
[1 +2 +\dots +k = \frac{k(k+1)}{2}]

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Step 3: Inductive Step ((n=k+1))

We need to show the formula holds for (n=k+1):

Compute the LHS for (n=k+1):
[1+2+\dots +k + (k+1)]

By the inductive hypothesis, substitute the sum up to (k):
[=\frac{k(k+1)}{2} + (k+1)]

Factor out ((k+1)):
[=(k+1)\left(\frac{k}{2} +1\right)]

Simplify the expression inside the parentheses:
[=(k+1)\cdot\frac{k+2}{2}]

[=\frac{(k+1)(k+2)}{2}]

This equals the RHS for (n=k+1): (\frac{(k+1)((k+1)+1)}{2}).

Thus, the formula holds for (n=k+1).

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Conclusion

By the principle of mathematical induction, the formula is true for all positive integers (n).

[ \boxed{1 +2 +\dots +n = \frac{n(n+1)}{2}} ]

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