is there a possible closed formula for the sequence 5,9,17,33,65,129..? because i cant find one... :(?

Answer 1

257 .... (129x2)-1

Answer 2

FORTHIS TYPE OF SEQUENCE U CAN ABLE TO ANSWER BY STUDYING IT ALONE
5
5+2^2= 5+4=9
9+2^3=9+8=17
17+2^4=17+16=33
33+2^5=33+32=65
65+2^6=65+64=129
THEREFORE 129+2^7=129+128=257
SO THE NEXT NUMBER IN THE SEQUENCE IS 257.

Answer 3

Prima facie, the value of each of successive terms is one less than twice the term. More specifically:
If M(r) denote the rth term of the sequence, then:
M(n) = 2*M(n+1) -1.

Now, to derive the terms M(r) in terms of r.
Write S(r) = M(1)+M(2)+....+M(r)
Then, S(r) = 5+9+17+33+65+129+......+M(r)
S(r) = 5+9 +17 + 33+ 65+ .....+ M(r-1) +M(r)
Subtracting, we obtain:
0 = 5+[ 4+8+.16+....+ 4*2^(r-2)] - M(r)
Or, M(r) = 5+ 4[2^(r-1) -1]
Or, M(r) = 2^(r+1) +1
This checks out since, M(1) = 5, M(2) = 9; M(3) = 17; M(4) = 33;
M(5) = 65; M(6) = 129 and, so on.
Thus; in general,
M(r) = 2^(r+1) +1 , which gives the rth term of the given sequence.

Answer 4

hayharbr's answer is correct. Many other answerers gave correct observations on the recurrence relationship, but I expect you needed the formula for the nth term without needing to work out the terms before it.

It's often worth looking at the pattern of differences. In this case the differences are 4, 8, 16, 32, .. and this gives the clue that powers of 2 are involved. It's then not very hard to observe that each term is just 1 more than a power of 2.

Answer 5

Start with 5

add 2^X to the previous term, where x is the term you are on.

so, 5, 5+4, 9+8, 17+16, 33+32, 65+64

Answer 6

I think what cdm0014 is asking is to find a formula f(n) where n is the n-th term
f(1) =5
f(2) =9
f(n) =2*f(n-1) -1

but find f(n) in which f(n-1) is not known.
Such as find the 20th term

f(1) = 5
f(2) = 5*2 -1
f(3) =5*2^2 -2 -1
f(4) =5*2^3 -2^2 -2 -1
f(5) =5*2^4 -2^3 -2^2 -2 -1
f(n) =5*2^(n-1) -2^(n-2) -2^(n-3) ...- 2 -1

This is a geometric series
-1(1 +2 +4 +...2^(n-2) ) +5*2^(n-1)

-1*(2^(n-1) -1) +5*2^(n-1)
-(2^(n-1) +1 +5*2^(n-1)
5*2^(n-1) + (-1)*2^(n-1) +1
4*2^(n-1) +1
(2^2)*2^(n-1) +1
2^(n+1) +1

f(n) = 2^(n+1) +1

Answer 7

It's increasing exponentially by 4.

5+4=9
9+4+4=17
17+4+4+4+4=33
33+4+4+4+4+4+4+4+4=65.....

or look at it like this

5+(4*1)=9
9+ (4*2)=17
17+ (4*4)=33

Answer 8

9-5=4
4x2=8 8+9=17

17-9=8
8x2=16 16+17=33

and so on...

Answer 9

x+(x-1)

Answer 10

Yes.
____5___4
____9___8
___17__16
___33__32
___65__64
__129_128

In BASIC:
10 A=5
20 B=4
25 Print A;
30 Print B
40 A=A+B
50 Print A;
60 B=2B
70 Print B
80 GOTO 40

;-D I love BASIC!

Answer 11

Here second term = 2 times the first term - one
Here third term = 2 times the second term - one
Here fourth term = 2 times the third term - one and so on......
Hence we can write
t(n+1) = t(n) -1 where n is greater then zero