Using Mathematical Induction Use mathematical induction to prove the formula for the sum of the first n squares. Sn = 1² + 2² + 3² + 4² + · · · + n² = n(n + 1)(2n + 1) / 6

Question

Using Mathematical Induction

Use mathematical induction to prove the formula for the sum of the first n squares.

[latex]S_n = 1^2 + 2^2 + 3^2 + 4 ^2 + \cdot \cdot \cdot + n^2 = \frac{n(n + 1)(2n + 1)}{6} [/latex]

Accepted Answer

1. When n = 1, the formula is valid, because

[latex]S_1 = 1^ 2 = \frac{1 (1 + 1)[2(1) + 1]}{6} = \frac{1(2)(3)}{6} = 1 [/latex].

2. Assuming the formula is true for k,

[latex]S_k = 1^ 2 + 2^2 + 3^2 + 4^2 + \cdot \cdot \cdot + k^2 = \frac{k (k + 1)(2k + 1)}{6} [/latex]

you must show that it is true for k + 1

[latex]S_{k+1} = \frac{(k + 1)[(k + 1) + 1][2(k + 1) + 1]}{6} = \frac{(k + 1)(k + 2)(2k + 3)}{6} [/latex].

To do this, write [latex]S_{k + 1}[/latex] as the sum of [latex]S_k[/latex] and the [latex](k + 1)[/latex] st term, [latex](k + 1)^{2} [/latex], as follows.

[latex] S_{k +1} = (1^2 + 2^2 + 3^2 + 4^2 + \cdot \cdot \cdot + k^2) + (k + 1)^{2} \\quad \quad = \frac{k(k + 1)(2k + 1)}{6} + (k + 1)^{2} \quad \quad \quad ext{ Induction hypothesis }\\quad \quad = \frac{(k + 1)(2k^2 + 7k +6)}{6} \quad \quad \quad \quad \quad \quad ext{ Combine fractions and simplify. } \\quad \quad = \frac{(k + 1)(k + 2)(2k +3)}{6} \quad \quad \quad \quad \quad\quad S_k ext{ implies } S_{k+1}. [/latex]

Combining the results of parts (1) and (2), you can conclude by mathematical induction that the formula is valid for all positive integers n.

Question Appendix.3: Using Mathematical Induction Use mathematical induction to p...

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