Question

Using the recursive formula, f(n+1)=2*f(n)+1, and given that f(1)=3, calculate the product of the digits in the number f(17).

Answer 1

what is f(2)?

Answer 2

\[f(1+1)=2~f(1)+1\]
\[f(2)=2(3)+1\]

knowing f(2), what is f(3)? and so on until you get to f(17)

Answer 3

I think you have to solve for f(n). You have the RHS = constant, that means you have to look for a f(Pn) = A
and then 2f(Pn) =2A
--------------------
f(Pn)) - 2 f(Pn) =-A =1 ---> A = -1
LHS you have the form of f(n+1) -2f (n) = 0 .The characteristic equation for that is
L(lamda) - 2 = 0
---> L =2
So the form of the function f(n) for the LHS is C *2^n
Now, combine both sides, f(n) =( C * 2^n) -1
when you have f(1) =3 replace to get C
(C*2) -1 =3
---> C = 2
Replace to form of f(n). You get the form f(n) = (2 * 2^n )-1
That is the final form to calculate the f of whatever value of n, It no longer depends on the previous term.
if you want to calculate f(17)? f(17) = (2 * 2^17) -1= 262143
Hope this what you need

Answer 4

@amistre64

Answer 5

ida put it in usual notation:\[a_n=2a_{n-1}+1~:~a_1=3\]
\[a_2=2(3)+1=7\]
\[a_3=2(7)+1=15\]
\[a_4=2(15)+1=31\]
\[a_n=2(31)+1=63\]

3 7 15 31 63
4 8 16 32

\[a_n=2^n-1\]

Answer 6

might have my index alittle outta sight

Answer 7

\[a_n=2^{n+1}-1\]might be better for me

Answer 8

which is what you have yes

Answer 9

you miss something, since a(n) = a(n -1) +1
so a (n+2) = a(n+1) +1 = a(n) +1 +1

Answer 10

thnks for medal.

Answer 11

ive seen your method about, but i have never had the time to actually learn it :)

Answer 12

I don't know, from my experience, I have to put everything in neat logic

Answer 13

so, you are genius in the field you never learn before, hat off

Answer 14

my teachers love me ;)

Answer 15

how lucky you are, my profs hate me most