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I have the Following Proof By Induction Question:

$$ (1)(2) + (2)(3) + (3)(4) + \cdots+ (n) (n+1) = \frac{(n)(n+1)(n+2)}{3} $$

Can Anybody Tell Me What I'm Missing.

This is where I've Gone So Far.

Show Truth for N = 1

LHS = (1) (2) = 2

RHS = $$ \frac{(1)(1+1)(1+2)}{3} $$

Which is Equal to 2

Assume N = K

$$ (1)(2) + (2)(3) + (3)(4) + \cdots+ (k) (k+1) = \frac{(k)(k+1)(k+2)}{3} $$

Proof that the equation is true for N = K + 1

$$ (1)(2) + (2)(3) + (3)(4) + \cdots+ (k) (k+1) + (k+1) (k + 2)$$

Which is Equal To: $$ \frac{(k)(k+1)(k+2)}{3} + (k+1) (k + 2)$$

This is where I've went so far

If I did the calculation right the Answer should be

$$\frac{(k+1)(k+2)(k+3)}{3}$$

Caleb Stanford
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Andrew
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2 Answers2

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Your proof is fine, but you should show clearly how you got to the last expression.

$\dfrac{k(k+1)(k+2)}{3}+(k+1)(k+2)$

$=\dfrac{k}{3}(k+1)(k+2)+(k+1)(k+2)$

$=(\dfrac{k}{3}+1)(k+1)(k+2)$

$=\dfrac{k+3}{3}(k+1)(k+2)$

$=\dfrac{(k+1)(k+2)(k+3)}{3}$.

You should also word your proof clearly. For example, you can say "Let $P(n)$ be the statement ... $P(1)$ is true ... Assume $P(k)$ is true for some positive integer $k$ ... then $P(k+1)$ is true ... hence $P(n)$ is true for all positive integers $n$".

Michael Hardy
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  • I know That the Final answer is $$\frac{(k+1)(k+2)(k+3)}{3}$$ by adding a k + 1 to every unknown from the Step 2 Which is Assume N = K$$\frac{(k)(k+1)(k+2)}{3}$$ – Andrew Feb 02 '13 at 13:19
  • Thanks for the Quick Edit, I only have one more question, What I'm Not understanding is how $$\frac{k(k+1)(k+2)}{3}$$ was changed to $$\frac{k}{3}+1$$ – Andrew Feb 02 '13 at 13:26
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It might also be helpful to go backwards to see that it makes sense. Specifically, $$S_{n+1} - S_n = \frac{(n+2)(n+1)(n)}{3} - \frac{(n+1)(n)(n-2)}{3} = \frac{1}{3}\left(n^3 + 3n^2 + 2n - n^3 + n\right) = n(n+1)$$ Going backwards, you can more easily see that $$S_{n+1} = S_n + n(n+1)$$

kaiwenw
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