1.15

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.

${\begin{aligned}S_{k+1}&=\sum _{i=1}^{k}i(i+1)(i+2)+a_{k+1}\\&={\frac {k(k+1)(k+2)(k+3)}{4}}+(k+1)((k+1)+1)((k+1)+2)\\&={\frac {k(k+1)(k+2)(k+3)}{4}}+(k+1)(k+2)(k+3)\\&={\frac {k(k+1)(k+2)(k+3)}{4}}+{\frac {4\cdot (k+1)(k+2)(k+3)}{4}}\\\end{aligned}}$

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.

Back to Chapter 1.