Divide each triangle into rows. Notice that each triangular number equals the sum of the number of points in each row.
C
c
Use mathematical induction. Add (k+1)(k+2)2 to both sides of the inductive hypothesis.
A
a
1, 3, 6, 10, 15, and 21
B
b
n(n+1)/2
C
c
See solution.
Practice makes perfect
a
According to the definition of a triangular number, we have that 3 is the second triangular number.
By adding three more points below the arrangement above we can get another triangular array. Thus, 6 is our third triangular number.
Now, to make another triangular arrangement, we can add 4 more points at the bottom of the latter triangle.
Following the pattern, we need to add 5 more points to the latter triangle.
Finally, we add a row with 6 points and get our next triangular number.
That way, we have found the first 6 triangular numbers.
1, 3, 6, 10, 15, 21
b
In Part A we found the first 6 triangular numbers.
1, 3, 6, 10, 15, 21
To write a formula that gives us the nth triangular number we will take a close look to the numbers above.
a_1 = 1, a_2 = 3, a_3 = 6
a_4 = 10, a_5= 15, a_6 = 21
From the above we cannot see a particular relation between the index and the term. But, remember that each triangular number is the number of points in the triangular array.
As we can see, a_n is the sum of the first n natural numbers.
a_n = 1 + 2 + 3 + ⋯ + n
Recall that the sum of the first n natural numbers can be found by computing n(n+1)2. This allows us to write a formula for the nth triangular number.
a_n = n(n+1)/2
c
In this part, we must prove that the sum of the first n triangular numbers equals the given fraction.
1 +3 +⋯+n(n+1)/2 = 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.
1( 1+1)( 1+2)/6 = 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 +⋯+k(k+1)/2 = 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 +⋯+( k+1)( k+1+1)/2
( k+1)( k+1+1)( k+1+2)/6
1 +3 +⋯+(k+1)(k+2)/2
(k+1)(k+2)(k+3)/6
To verify that both sides are equal, we will start with the inductive hypothesis and perform some operations.
