Question

Find a closed form for the following sequence and the next 2 terms: n>= 0

3, 7, 16, 35, 74, 153, 312, 631, 1270, ....

Answer 1

2549 , 5108

Answer 2

lol, apart from the wolf

Answer 3

\[n_0,2n_0+1,2n_1+2,2n_2+3,...\]seems like the pattern, I don't know how to write it generally.

Answer 4

3
7 = 3*2 +1
16 = 7*2 + 2
35 = 16*2 +3
74 = 35*2 +4
153 = 74*2 + 5
312 = 153*2 +6
631 = 312*2 + 7
1270 = 631*2 + 8

Answer 5

\[\left\{ n_i \right\}=2n_{i-1}+i\]i think right?

Answer 6

that might be a good recursive rule for it

Answer 7

from with \[n_0=3\]

Answer 8

its pattern is 2n(i-1) +(i-1)

Answer 9

@ amistre i nvr take help of Wolf

Answer 10

..... i do lol

Answer 11

u commented something wolf above

Answer 12

ive been working on a way to see if I can find a pattern to a sequence that has differences that can be brought down to a row that is a multiple of a constant ... if that makes sense

Answer 13

if you take 2 differences the 3rd row is 5(2^n) on this one

Answer 14

\[d_1+n(d_2-\frac{5}{2-1})+5\frac{2^n-1}{(2-1)^1}\]
is what i got so far

Answer 15

that last terms should be (2-1)^2

Answer 16

I frankly don't work with sequences enough to help on this one, sorry :(

Answer 17

is ok, its self generated ;)

Answer 18

a better expression would be to say , the sequence above is of the form, a0, a1, a2, ...

\[a_0+n(d_1-\frac{d_2}{r-1})+d_2\frac{r^n-1}{(r-1)^1}\]
where d1 is start of the row of the 1st differences

and d2 is start of the row of the 2nd differences

and r is the common ratio associated with the 2nd row.

r0: 3 7 16 35 74 153
r1: 4 9 19 39 79 ; thats seems patternish.
r2: 5 5(2) 5(4) 5(8)

\[r_1=4+5\sum2^n\]

\[r_0=3+\sum(4+5\sum2^n)\]
\[r_0=3+4n+\frac{5}{2-1}\sum(2^n-1)\]