# MAT helpWatch

#1
Hey Guys

Just going through all the MAT papers again, and I came across a question which bothered me not because I couldn't get an answer but I did not have a sure fire method of getting an answer I got the marks but it was due to abit of trial and error.

Fn(x)=(2+(-2)^n)x^2+(n+3)x+n^2

Now F1(x)=4x+1

F(F1(x))=16x+5

F(F(F1(x)))= 64x+21

Find an expression for where f1 is applied k times.

Well it's pretty obvious for the x term it shall be (4^k)x. However the constants are a little more tricky as they do not have a constant common ratio or difference but I noticed the difference of the difference does I played around with this and I got ((4^k)-1)/3 but there must a more solid approach to this question thanks in advance.
0
4 years ago
#2
When you want to think about something like this, write it out. e.g.

expanded out, gives constant term 1 + 4 + 4^2 + ....
This is a geometric series. I'm sure you can figure out the rest.
1
#3
Ahh this is a a great way to look at it thanks for the help
0
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