how would i derive a formula for the sequence [math]a_{n}[/math]?
>>7687747
I could not help but notice your png was not optimized anon.
I have optimized your png.
Your png is now optimized.
>>7687747
You just shift all n's to the left by 1.
[math]a_n = a_1 + a_2 + ... + a_{n-1}[/math]
The reason you don't do this is because the equation doesn't work when n = 1 (what's [math]a_0[/math]?). You'd have to specify that the equation only works at n > 1 which makes it more complicated to understand.
If you take [math] a_0 = 1 [/math] then [math] a_n = 2^{n-1}[/math] for [math] n\geq 1 [/math].
>>7687747
>retarded CS major detected
[math] a_{ n+1 } = \sum _ { i=1 } ^ { n } a_i = \sum _ { i=1 } ^ { n-1 } a_i + a_n = 2*a_n [/math]
[math] a_n = ⌈ 2 ^ { n-2 } ⌉ [/math]
>>7687785
> ceiling function
>>7687773
>>7687773
>>7687815
>>7687815
>>7687817
Only doing lossless compression. Though that's more or less visually transparent.
Did you use some sort of color quantization?
>>7687812
n=1
>>7687815
>>7687747
Notice that for n > 1, you have [math]a_n = a_1 + \dots + a_{n-1}[/math]
Therefore [math]a_{n+1} = 2a_n [/math] for [math]n \ge 2[/math]
Since [math]a_2 = 1[/math], you get [math]a_n = 2^{n-2}[/math] for [math]n \ge 2[/math]
>>7687747
if you look closely, you can see that [math]a_{n+1} = a_n + a_n = 2a_n [/math] (the first n terms add up to an)
