Calculate the asymptotic growth of a sum that contains log or binom

Question

I'm looking for a basic explanation how to calculate the asymptotic growth of sums.

Take for example this one:

$\sum_{i=1}^{lg(n!)} 2^{n^2}$

or this one:

$\sum_{i=0}^{n} {n\choose{i}}$

The solition of the first one is: $lg(n!)2^{n^2}$ and for the second one is: $2^n$

note: this is NOT a homework, The above tasks are sample problems for the DAA's exam and I will be really grateful if someone spare some time and explain it for me.

Answer 1

Notice that in the first sum the summand, $2^{n^2}$, does not depend on the summation index, so $\sum_{i=1}^{\log (n!)} 2^{n^2} = 2^{n^2}\sum_{i=1}^{\log (n!)} 1 = 2^{n^2}\log(n!)$.

For the second sum, notice that by the Binomial theorem we have $2^n = (1+1)^n = \sum_{i=0}^n 1^i1^{n-i}\binom{n}{i} = \sum_{i=0}^n\binom{n}{i}$.