Proof by Induction: The Sum of Squares Formula

Theorem. For any integer n ≥ 1,

∑_i=1ⁿ i² = (1/6)·n·(n+1)·(2n+1)

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Proof (by mathematical induction)

1. Basis Step

For n = 1:

LHS = 1² = 1
RHS = (1/6)·1·(1+1)·(2·1+1)
    = (1/6)·1·2·3
    = 1

Therefore, the formula holds for n = 1.

2. Inductive Hypothesis

Assume the statement is true for some integer k ≥ 1:

∑(i=1→k) i² = (1/6)·k·(k+1)·(2k+1)

3. Inductive Step

We must show it is true for n = k + 1:

∑(i=1→k+1) i²
= [∑(i=1→k) i²] + (k+1)²
= (1/6)·k·(k+1)·(2k+1) + (k+1)²
= (1/6)(k+1)[k(2k+1) + 6(k+1)]
= (1/6)(k+1)[2k² + 7k + 6]
= (1/6)(k+1)(k+2)(2k+3)

This matches the same form with k replaced by k+1:

(1/6)·(k+1)·((k+1)+1)·(2(k+1)+1)

Hence, the formula holds for n = k + 1.

4. Conclusion

Since the statement is true for n = 1 (basis step), and true for n = k + 1 whenever it is true for n = k (inductive step), it follows by the principle of mathematical induction that

∑(i=1→n) i² = (1/6)·n·(n+1)·(2n+1)

for all integers n ≥ 1.

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Notes

• This summation identity plays a key role in deriving the Riemann sum for ∫₀ˣ t² dt = (1/3)x³.
• The simplification 2k² + 7k + 6 = (2k + 3)(k + 2) is the key algebraic step.
• Mathematical induction demonstrates how a result true for one case extends to all cases.

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