There is no unique formula. Given a set of any 7 numbers it is always possible to find a polynomial of order 6 that will generate those numbers as the first set of numbers. For example,
Un = (n6 - 33n5 + 475n4 - 3915n3 + 20044n2 - 61932n + 91440)/360
for n = 1, 2, ... , 7 will give the above 7 numbers.
You need to be able to second-guess what the questioner had in mind. In this case you may notice that each number is half the previous one. So the expected answer is
Un = 28-n.
Mathematically, both answers are correct but unfortunately, in most cases the first will not be accepted.
1, 2, 4, 8, 16, 32, 64, 128.
11000000 = 128+64=192 10101000 = 128+32+8=168 10101100 = 128+32+8+4=172 11110001 = 128+64+32+16+1=241
128 = 2 x 64 = 4 x 32 =8 x 16
The GCF is 32.
The positive-integer factors of 128 are: 1, 2, 4, 8, 16, 32, 64, and 128
1, 2, 4, 8, 16, 32, 64, 128, -1, -2, -4, -8, -16, -32, -64, -128
32, 64, 96, 128, 160, 192, 224, 256, 288, . . .16, 32, 48, 64, 80, 96, 112, 128, 144, . . .
1, 2, 4, 8, 16, 32, 64, 128: 1 x 128, 2 x 64, 4 x 32, 8 x 16
1, 2, 4, 8, 16, 32, 64, 128.
The factors of 128 are: 1, 2, 4, 8, 16, 32, 64, 128.
1, 2, 4, 8, 16, 32 1, 2, 4, 8, 16, 32, 64, 128 1, 2, 4, 8, 16, 32, 64, 128, 251, 502, 1004, 2008, 4016, 8032, 16064, 32128
1 x 128, 2 x 64, 4 x 32, 8 x 16, 16 x 8, 32 x 4, 64 x 2, 128 x 1 = 128
512 / 2 = 256 256 / 2 = 128 128 / 2 = 64 64 / 2 = 32 32 / 2 = 16 16 / 2 = 8 8 / 2 = 4 4 / 2 = 2 2 / 2 = 1
1, 2, 4, 8, 16, 32, 64, 128.
By: 1 2 4 8 16 32 64 and 128
1, 2, 4, 8, 16, 32, 64, 128.
1, 2, 4, 8, 16, 32, 64, 128.