1
$\begingroup$

I need to know which sequence grows faster with n:

$$ f(n) = \sum_0^{floor(n/3)} {n \choose 3*i+1} $$

compared to

$$ g(n) = 2^n -1 $$

it seems f(10)>5000 is greater than g(10)=1023 but I would like to know what happen for greater n's

edit: is $f(n)$ equivalent $n^4$ so the g(n) should grow faster?

edit2: sorry I forgot $\sum_0^n {n \choose i} = 2^n$

  • 1
    What is $E$? If it's less than $1$, explicit formulas seem out of the picture.2012-05-30
  • 0
    it's round(x)-12012-05-30
  • 0
    How do you get $f(10)>5000$ when $f(n)$ is certainly less than the sum of all the coefficients on the $n$-th, which is $2^n$ ??2012-05-30
  • 0
    yes wrong calcul2012-05-30

2 Answers 2