Does this series $2 + 4 + \cdots + \sqrt{\sqrt{n}} + \sqrt{n} + n$ have a general term?

Question

Does this sum simplify to a general term in terms of $n$? If so, how would you arrive at that term?

$2 + 4 + \cdots + \sqrt{\sqrt{n}} + \sqrt{n} + n$.

Thanks.

Answer 1

As noticed by Semsem, there is an inconsistency between the first three terms and the general term.

Let us assume that you have $$a_i= \sqrt[4]{i}+\sqrt{i}+i$$ and that you ask for the summation of the $a_i$ for $i$ from $1$ to $n$, there is a formula for $$\sum _{i=1}^n a_i=\frac{1}{2} n (n+1)+H_n^{\left(-\frac{1}{2}\right)}+H_n^{\left(-\frac{1}{4}\right)}$$ If the summation starts at $i=2$ (as it seems in the post), then $$\sum _{i=2}^n a_i=\frac{1}{2} (n-2) (n+3)+H_n^{\left(-\frac{1}{2}\right)}+H_n^{\left(-\frac{1}{4}\right)}$$ where appear the harmonic numbers.

Answer 2

By looking at the general terms you can see that each number is the square of its previous. Is this what you're looking for? $$f(k)=2^{2^k}$$

This one assumes that your first number is incorrect. So your sequence is: $2, 4, 16, 256...$