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# How do I get to this answe?

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1. Let fn(x)=(2 + (−2)n ) x2 + (n+3)x + n2 where n is a positive integer and x is any real number.
Find an expression, simplified as much as possible, for f1 (f1 (f1 (··· f1 (x)))) where f1 is applied k times.
[Here k is a positive integer.]

More generally f(1 k times)(x) = 4k x + 1 + 4 + · · · + 4k−1= 4k x +(4^k − 1)/3

How do you work this answer out (The '(4^k -1)/3' part of the answer).

Basically, how do you get from (4k-1 + 4k-2 + ... + 4k-k)

to (4^k -1)/3
2. (Original post by KloppOClock)
Let fn(x)=(2 + (−2)n ) x2 + (n+3)x + n2 where n is a positive integer and x is any real number.
Find an expression, simplified as much as possible, for f1 (f1 (f1 (··· f1 (x)))) where f1 is applied k times.
[Here k is a positive integer.]

More generally f(1 k times)(x) = 4k x + 1 + 4 + · · · + 4k−1= 4k x +(4^k − 1)/3

How do you work this answer out (The '(4^k -1)/3' part of the answer).

Basically, how do you get from (4k-1 + 4k-2 + ... + 4k-k)

to (4^k -1)/3
Are you sure it's not instead??
3. (Original post by RDKGames)
Are you sure it's not instead??
https://www.maths.ox.ac.uk/system/fi...olutions07.pdf

question 2 part ii
4. Ah I see, never mind, was thinking slightly differently.

Okay you have:

This is a geometric sequence and you need to use a formula to sum it up where where a is the first term and r is the common ratio.
5. (Original post by RDKGames)
Ah I see, never mind, was thinking slightly differently.

Okay you have:

This is a geometric sequence and you need to use a formula to sum it up where where a is the first term and r is the common ratio.
ahh right, i forgot all about sequences and series. thank you.

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