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Call the statement $S_n$ and the general term $a_n$

Step 1: Show that the statement holds for the basis case $n = 0$

$a_0 = 0 \cdot (0 + 1)(0 + 2) = 0$

$S_0 = \frac {0 \cdot (0 + 1)(0 + 2)(0 + 3)} {4} = \frac {0} {4} = 0$

Since $a_n = S_n$, the basis case is true.

Step 2: Assume that $n = k$ holds.

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

Step 3: Show that on the assumption that the summation is true for k, it follows that it is true for k + 1.

\begin{align} S_{k + 1} & = \sum_{i = 1}^k i(i + 1)(i + 2) + a_{k + 1} \\ & = \frac{k(k + 1)(k + 2)(k + 3)} {4} + (k + 1)((k + 1) + 1)((k + 1) + 2) \\ & = \frac{k(k + 1)(k + 2)(k + 3)} {4} + (k + 1)(k + 2)(k + 3) \\ & = \frac{k(k + 1)(k + 2)(k + 3)}{4} + \frac{4 \cdot (k + 1)(k + 2)(k + 3)} {4} \\ \end{align}

It's easier to factor than expand. Notice the common factor of (k + 1)(k + 2)(k + 3).

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

This should be equal to the formula $S_k = \frac{k(k + 1)(k + 2)(k + 3)}{4}$ when k = k + 1:

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

Since both the basis and the inductive step have been proved, it has now been proved by mathematical induction that S_n holds for all natural n