`a_n=(-2/5)^n`
Assume n begins with 1,
`a_1=(-2/5)^1=-2/5`
`a_2=(-2/5)^2=(-2/5)(-2/5)=4/25`
`a_3=(-2/5)^3=(-2/5)(-2/5)(-2/5)=-8/125`
`a_4=(-2/5)^4=(-2/5)(-2/5)(-2/5)(-2/5)=16/625`
`a_5=(-2/5)^5=(-2/5)(-2/5)(-2/5)(-2/5)(-2/5)=-32/3125`
The first five terms of the sequence are `-2/5,4/25,-8/125,16/625,-32/3125`
Comments
Post a Comment