Question 1017792
A sequence {m<sub>n</sub>} is recursively defined as 
m<sub>1</sub>=2, m<sub>n+1</sub>=3/5-m<sub>n</sub>
<pre>

substitute n=1
m<sub>1+1</sub>=3/5-m<sub>1</sub>
m<sub>2</sub>=3/5-2 = 3/5-10/5 = -7/5

substitute n=2
m<sub>2+1</sub>=3/5-m<sub>2</sub>
m<sub>3</sub>=3/5-(-7/5) = 3/5+7/5 = 10/5 = 2

Wow! We're right back where we started.  If we continue
with this recursion formula:

substitute n=3
m<sub>3+1</sub>=3/5-m<sub>3</sub>
m<sub>4</sub>=3/5-2 = 3/5-10/5 = -7/5

substitute n=4
m<sub>4+1</sub>=3/5-m<sub>4</sub>
m<sub>5</sub>=3/5-(-7/5) = 3/5+7/5 = 10/5 = 2

So the sequence goes:

2, -7/5, 2, -7,5, 2, -7/5, ...

Edwin</pre>