2007-11-10 15:33:26 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
7, 11, 19, 35, 67, 131, 259, 515, 1027, 2051, 4099...
Double, then subtract 3
7 x 2 - 3 = 11
11 x 2 - 3 = 19
19 x 2 - 3 = 35
35 x 2 - 3 = 67
etc.
Another way to calculate each term is:
a(n) = 2^(n+1) + 3
a(1) = 2^2 + 3
a(1) = 4 + 3
a(1) = 7
a(2) = 2^3 + 3
a(2) = 8 + 3
a(2) = 11
a(3) = 2^4 + 3
a(3) = 16 + 3
a(3) = 19
a(4) = 2^5 + 3
a(4) = 32 + 3
a(4) = 35
a(5) = 2^6 + 3
a(5) = 64 + 3
a(5) = 67
2007-11-10 15:37:29 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
the next number in series is 67
(let us assume that the number required is x .
finding the difference between the no ie.11-7 = 4
19-11=8
35-19=16
x-35=32
x=35+32
x=67
(the right hand side value is just getting double i.e 4,8,16,the next number to be taken is 32)
2007-11-10 15:43:39 · answer #2 · answered by rrr 2 · 0⤊ 0⤋
11
2007-11-10 15:41:04 · answer #3 · answered by Anonymous · 0⤊ 0⤋
67
2007-11-10 15:41:32 · answer #4 · answered by imbustass 4 · 0⤊ 0⤋
7,11,19,35,67
Add 4, 8 16, 32, 64 and so on.
2007-11-10 15:42:51 · answer #5 · answered by vaderismydog2006 3 · 1⤊ 0⤋
7,11,19, 35, 67,...
a(1) = 7
a(n+1) = a(n)+2^(n+1)
2007-11-10 15:43:18 · answer #6 · answered by sahsjing 7 · 0⤊ 0⤋