Why Is The Manual Summation Of An n log n Equation Not Equal To The Programmatic Summation Of An θ(n log n) for Loop?

There is an algorithm text book that I'm reading to teach myself asymptotic analysis. To demonstrate that not all two-level nested for loops are $\Theta(n^2)$, the book presents the following code to be analyzed...

/* θ(n log n) */
sum = 0;
for(int i = 1; i <=n; i*= 2)
    for(int j = 1; j <=n; j++)
        sum++;

The book calculated that the running-time of the above implementation would be $\Theta(n \log n)$. The book arrived at the $\Theta(n \log n)$ running-time, after first suggesting that the intermediate Sigma notation is $\sum_{i=1}^{\log n} n = n \log n$.

Before I ask my question, I'll first set up the context of my question by showing another block of code with $\Theta(n^2)$ running-time — whose corresponding Sigma notation is $\sum_{i=1}^{n} i = \frac{n(n-1)}{2}$...

/* θ(n^2) */ 
sum = 0;
for(int i = 0; i <=n-1; i++)
    for(int j = n-1; j > i; j--)
        sum++;

If the above-listed $\Theta(n^2)$ code were run with $n = 8$, then the value of the variable sum, would end up being 36. Correspondingly, the manual calculation of the $\Theta(n^2)$ closed-form summation equation ($\frac{n(n-1)}{2}$) would work out to be 36 if you were to plug the value 8 into its $n$.

If — on the other hand — the above listed $\Theta(n \log n)$ code were run with $n = 8$, then the value of its sum variable would end up being 32. However, the manual calculation of $\Theta(n \log n)$ — where $n = 8$ — would work out to be 24. Because asymptotic analysis always assumes $log_2$, then $\log 8 = 3$. And $8 \cdot 3 = 24$.

So now, here are my questions:

1. How come the $\Theta(n^2)$ code's summation and the summation of its corresponding Sigma notation's closed form equation both work out to be 36 — but the $\Theta(n \log n)$ Sigma notation equation calculation does not jibe with the value calculated by the corresponding $\Theta(n \log n)$ code listed above?

2. What is the relationship between the 24 that comes out of the $\Theta(n \log n)$ Sigma notation equation calculation and the corresponding code listed above?

3. What would I need to change in the above-listed $\Theta(n \log n)$ code, to make the final value of the summation, equal to the summation calculated in its corresponding Sigma notation equation?

4. What would the the $\Theta(n \log n)$ Sigma's closed-form equation ($n \log n$) look like fully expanded — where $n = 8$?

Thank you in advance for your answers.

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EDIT: Please, don't let my use of MathJax fool you into thinking I'm some kind of Math dude. 'Coz I'm not. I suck at Math. Which is the very reason why I couldn't answer the question on my own. I also suck at Math jargon. Therefore, I would not be offended in the least if any answers and comments used ELI5-level plain English. In fact, I'm gonna have to insist on jargon-less, ELI5-level plain English. Please? Thanks.

Which one of the four (count 'em) questions in my OP — five counting the one in the title — confuses you? Because you've addressed neither of them.

@algoHolic - $n \cdot (\log(n) + 1)$ with $n=8$ will give you 32. And please be creative, but not aggressive

I wasn't being creative. I was being repetitive. Meaning: I only repeated your own usage of "confused", @HEKTO. Nor was I being "creative" with the $\sum_{i=1}^{\log n} n = n \log n$ Sigma notation. Like I've already said in my OP — I'm only repeating the Sigma notation that I copied from an algorithm analysis text book I'm reading. Like I also said — The book based its $\Theta(n \log n)$ running-time solution on what the book claims was the summation of the closed-form of the $\sum_{i=1}^{\log n} n = n \log n$ Sigma notation that the book presented to support its solution...

...There are dozens of algorithm-related, Sigma notated summations that contain constant terms in their closed-form equations. The above-mentioned $\sum_{i=1}^{n} i = \frac{n(n-1)}{2}$ is one example. So, as a naive self-learner, I gotta figure that the author intentionally presented his $\sum_{i=1}^{\log n} n = n \log n$ summation for some specific reason. Which begs the question: What IS the reason the author uses $\sum_{i=1}^{\log n} n = n \log n$ to represent the summation of the above $\Theta(n \log n)$ code? — (that I copy/pasta-ed from the book, I hasten to repeat)...

...If $\sum_{i=1}^{\log n} n = n \log(n) + 1$ was indeed what the author intended, there is no obvious reason why he would not have just tacked the stupid $+ 1$ onto the end of his equation. After all, by that point in the book, he'd already drilled the universally parotted, "Big-O ignores constants and low-order terms" dogma into the reader. So it would have been abundantly clear to me how to deal with the $+ 1$ term in the Sigma notation, if he hadda included it. Granted — maybe the author himself is confused.

I didn't create these either. Cormen, et. al did... $$\sum_{k=0}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$ $$\sum_{k=0}^n k^3 = \frac{n^2(n+1)^2}{4}$$ $$\sum_{k=0}^{n} x^k = \frac{x^{n+1} - 1}{x - 1}$$ $$\sum_{k=0}^\infty x^k = \frac{1}{1-x}$$ Notice how all the closed-form equations have constant terms?

Just curious: I am clueless as to what "ELI5-level" means. Seriously. I'm from the US, so I'm guessing it relates to a specific level (grade-level) of mathematics. But if you could be so kind as to educate my guesses, I'd be much obliged.

To answer your Q @amWhy : "•E is for Explain - merely answering a question is not enough" | "•LI5 means friendly, simplified and layman-accessible explanations - not responses aimed at literal five-year-olds" | [Source] | It comes from an old adage (and I paraphrase) — "If you can't explain something simple and straightforward enough so that a ten year old would understand it — then you really don't understand that something yourself"...

Thanks for the lesson! :-)