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Using the Principle of Mathematical Induction, prove that . n ^2 − n is even for n ≥ 1​

User RGil
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Base case: if n = 1, then

1² - 1 = 0

which is even.

Induction hypothesis: assume the statement is true for n = k, namely that k ² - k is even. This means that k ² - k = 2m for some integer m.

Induction step: show that the assumption implies (k + 1)² - (k + 1) is also even. We have

(k + 1)² - (k + 1) = k ² + 2k + 1 - k - 1

… = (k ² - k) + 2k

… = 2m + 2k

… = 2 (m + k)

which is clearly even. QED

User Akrohit
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