Question

]What is the rule for the sequence 13, 26, 52, 104, …?

Answer 1

i can come up with "a" rule for it that will hit at least those numbers

Answer 2

```
f(n)=a(n-2)(n-3)(n-4) + b(n-1)(n-3)(n-4) + c(n-1)(n-2)(n-4) + d(n-1)(n-2)(n-3)
```

Answer 3

ugh, i added instead of subtracted .... silly me

Answer 4

\[13 = a(-1*-2*-3);a=-13/6\]\[26 = b(1*-1*-2);b=13\]\[52 = c(2*1*-1);c=-26\]\[104 = d(3*2*1);d=104/6\]
y=-13(n-2)(n-3)(n-4)/6 + 13(n-1)(n-3)(n-4) -26(n-1)(n-2)(n-4) + 104(n-1)(n-2)(n-3)/6
or expanded as\[y=\frac{13}6n^3-\frac{13}2n^2+\frac{52}3n\]

Answer 5

but the pattern is just to double the number before to get the next one

Answer 6

an = 13*2^(n-1)