a Note that T_n can be written as the finite series of an arithmetic sequence.
B
b Use the explicit rules from Part A.
C
c Note that a square S_n can be divided into two smaller triangles, T_n and T_(n-1).
A
a T_n=n(n+1)/2, S_n=n^2
B
b T_1=1, T_n=T_(n-1)+n for n>1 S_1=1, S_n=S_(n-1)+2n-1 for n>1
C
c S_1=T_1 and S_n=T_(n-1)+T_n for n>1
Practice makes perfect
a First, let's analyze the sequence T_n.
Triangular Numbers
The numbers T_n are represented by the points of following diagrams.
Note that the n^(th) triangular number, T_n, is the sum 1+2+...+n. We will find the explicit rule using the formula for finding the sum of an arithmetic series.
T_n=n( a_1+ a_n)/2
For this series, the first term is a_1= 1 and the n^(th) term is a_n= n. Now, let's substitute the values and find the explicit rule for T_n.
Therefore, the explicit rule for the sequence of triangular numbers is T_n= n(n+1)2.
Square Numbers
The n^(th) square number, S_n, represents the number of dots in a square with the side made of n dots. Therefore, the total number of dots is n* n=n^2. This tells us that the explicit rule for the sequence of square numbers is S_n=n^2.
b Of course, the first triangular number is T_1=1.
Notice that the number of dots in T_n is n more than the number of dots in T_(n-1). Therefore, we can write a recursive rule for T_n.
T_1=1
T_n=T_(n-1)+nforn>1
Now, let's deal with the square numbers. The first square number is S_1=1.
Notice that the number of dots in S_n is (n-1)+(n-1)+1=2n-1 more than the number of dots in S_(n-1). Therefore, we can write a recursive rule for S_n.
S_1=1
S_n=S_(n-1)+2n-1forn>1
c Of course T_1=S_1. Now, let's analyze the situation when n>1. Notice that a square S_n can be divide into two smaller triangles, T_n and T_(n-1), as in the following picture.
