My favorite 3 concepts in 5 problems!

I don't like deleting backslahes...

Solve each problem one by one...

1) Note that you can factor two perfect squares like (a+b)(a-b) from the form $a^2-b^2$ . So if we do this, we get (3999)(1), equaling $\boxed{3999}$

2) Remember Gauss? Yep, he did this problem. For those who don't understand how Gauss as a little child did it so fast, he figured out:

$1+2+3+...+98+99+100$

Made pairs so that

$1+100 = 101, 2+99 = 101, 3+98 = 101,....$

Since there are 50 pairs like this, we get $50 \cdot 101$ , which equates to $\boxed{5050}$ .

3) Do what Gauss did again. This time, the pairs equate to 201, and there are 100 of them. Thus, $100 \cdot 201$ = the answer, $\boxed{20100}$

4) Remember Binomial Theorem! $\binom{4}{3}$ will give you the coefficient of the term, which equals to $\frac{4!}{3!}$ = $\boxed{4}$

5) Now here, I smashed all my favorite concepts into one little problem. Again:

Remember Binomial Theorem!

Again, $\binom{7}{5}$ gives you $\frac{7!}{5!2!}$ , which gives $\frac{7\cdot6}{2}$ , making N equal to $21$ .

Factor the squares. $21^2-20^2 = (21-20)(21+20) = 41 \cdot 1$ as M. Now that M equals 41, we are asked to find the first 41 positive integers.

I like to have an even number of positive integers. (how can you have 20.5 pairs of integers?) so lets find the first 40 positive integers and add 41 to it. Now we have the pairs equating to 41 (oh wow) and there are 20 of them. $41\cdot20 = 820$ . Adding 41 to it should give you $\boxed{861}$

Now, we (as Raj Magesh said, and I can't say it any better):

Concatenating the answers, we get:

Now so, concatenating the answers, we get $\boxed{39995050201004861}$