c In this part, we must prove that the sum of the first

n

triangular numbers equals the given fraction.

1 + 3 + \dots + \frac{n ( n + 1 )}{2} = \frac{n ( n + 1 ) ( n + 2 )}{6}

To prove this we will use mathematical induction.

Step $1$

The first step is to show that the given equation is true for

n = 1 .

When

n = 1,

the left-hand side becomes

1 .

Let's check what happens to the right-hand side.

\frac{1 ( 1 + 1 ) ( 1 + 2 )}{6} = \frac{1 ( 2 ) ( 3 )}{6} = 1

As we can see, the right-hand side equals

1

as well. Thus, the given equation is true for

n = 1 .

Step $2$

In this part, we will settle down the inductive hypothesis. That is, we will assume that the given equation is true for a natural

k .

1 + 3 + \dots + \frac{k ( k + 1 )}{2} = \frac{k ( k + 1 ) ( k + 2 )}{6}

Step $3$

Here, we have to show that the statement is true for $n = k + 1 .$ Let's see how the left and right-hand sides change when we substitute $n = k + 1 .$

Left-hand side	Right-hand side
$1 + 3 + \dots + \frac{( k + 1 ) ( k + 1 + 1 )}{2}$	$\frac{( k + 1 ) ( k + 1 + 1 ) ( k + 1 + 2 )}{6}$
$1 + 3 + \dots + \frac{( k + 1 ) ( k + 2 )}{2}$	$\frac{( k + 1 ) ( k + 2 ) ( k + 3 )}{6}$

Exercise