# Difference between revisions of "TADM2E 1.13"

Call the statement $S_n$ and the general term $a_n$<br>

<b>Step 1:</b> Show that the statement holds for the basis case $n = 0$<br>

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

<br><br>

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

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

<b>Step 2:</b> Assume that $n = k$ holds.

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

<b>Step 3:</b> 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}<br>

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

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

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

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

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