Calculate S = [1 / √2 + √1] + [1 / √3 + √2] + … + [1 / √100 + √99].
Determine the smallest natural number n such that: [1 / √2 + √1] + [1 / √3 + √2] + … + [1 /
√n+1 + √n].
1. We can do it with the calculator but it will take too much time. Is there a faster way to do it?

2.[1 / √2 + √1] + [1 / √3 + √2] + … + [1 / √n+1 + √n]
= (√2+√1) + (√3+√2) + … + (√n+1 + √n)
= √n+1 – 1

so finding the first n such that the sum in question exceeds 100 amounts to solving:
√n+1 – 1 ≥ 100
⇔ √n+1 ≥ 101
⇔ (√n+1)² ≥ 101²
⇔ n+1 ≥ 10201
⇔ n ≥ 10200

Is it good ?

And actually 2+2=4...
U mean 2+20?

An other problem :
( 0.8 X (e^-0.2t)) / ( 1+ 4X(e^-0.2t))² > 0.9

I started by putting 0.9 on the same denominator and I get this:

(0.8e^-0.2t) > 0.9 ( 1 + 4e^-0.2t)²

and there, it becomes more complicated...I tried to develop the left parenthesis (remarkable identity) but I end up with a (4e^-0.2t)² which bothers me enormously.

Try figure this out >:)

2 + 2 = 22
How?

Lol so easy XD
it's a calculation of primary this!