# 1.15

Call the statement $\displaystyle{ S_n }$ and the general term $\displaystyle{ a_n }$

Step 1: Show that the statement holds for the basis case $\displaystyle{ n = 0 }$

$\displaystyle{ a_0 = 0 \cdot (0 + 1)(0 + 2) = 0 }$

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

Since $\displaystyle{ a_n = S_n }$, the basis case is true.

Step 2: Assume that $\displaystyle{ n = k }$ holds.

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

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

\displaystyle{ \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).

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

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

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

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.

Back to Chapter 1.