TADM2E 1.13

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Call the statement <math>S_n</math> and the general term <math>a_n</math><br>

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

<math>a_0 = 0 \cdot (0 + 1)(0 + 2) = 0</math>

<br><br>

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

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

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

<math>S_k = \frac {k(k + 1)(k + 2)(k + 3)} {4}</math><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.

<math>

\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} </math><br>

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

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

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

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

\frac{(k + 1)(k + 2)(k + 3)(k + 4)} {4}</math><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

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