Ehrlich

The answer is 2

Whats the point of your first question? couldnt you just call the monstrosity on the right side of the equation 'x'?

and I didnt really get your second question.

\(\lim_{n\rightarrow infinity}{{(\sqrt[n]{|{(-1)}^{n}*(1-(\frac{15}{n}))|})}^{n}}^{2}= \lim_{n\rightarrow infinity}{|{(-1)}^{n}*(1-(\frac{15}{n}))|}^{2}=(\lim_{n\rightarrow infinity}{|{(-1)}^{n}*(1-(\frac{15}{n}))|})^2=1^2=1.\)

But the function is 4/e*e^x-x-1 isnt it? isnt it f(2)=4e-3?

There is also a generalization of this question:

aⁿ+bⁿ/(ab)^n-1+1 is an integer k. Prove that k=cⁿ where c is also an integer (a, b and n are all integers)

"your definately not 4."

what?

I dont think finding f(2) is possible....

f'(x)=x+f(x)--------->

f'''(x)=(x)''+f''(x)=f''(x)----->

f''(x)=c₁*e^x------->

f(x)=c₁*e^x+c₂*x+c₃

f(1)=c₁*e+c₂+c₃=2-------------->

c₃=2-c₂-c₁*e

f(2)=c₁*e²+c₂*2+c₃=

c₁*e²+c₂+2-c₁*e

edit: OH SHOOT I FORGOT SOMETHING

f'(x)=c₂+c₁*e^x=x+c₃+c₂*x+c₁*e^x  | subtract c₁*e^x ----->

c₂=x+c₃+c₂*x. therefore, c₂=-1(the only option. that means c₃=c₂=-1 and that means

f(1)=c₁*e-2=2 therefore c₁=4/e

that means f(2)=4*e-2-1=4e-3.

I didnt understand the guest's way, so ill solve it my way:

P(k)=the number of times the digit k appears between 0 and 999,999. P(0)*0+P(1)*1+P(2)*2+....+P(9)*9+1= the sum we need to find (i replaced 1,000,000 with 0, all we need to do is add 1 to the final solution.

all the numbers between 0 and 999,999 are also every possible combination of 6 digits (instead of 1 we write 0000001, instead of 2 we write 000002. Yes, a large number of zeroes is added, but that doesnt change the solution because x+n*0=x+0=x)

Here's a link to a similar question someone asked about the number of times a digit appears between 1 and 10ⁿ(n is a positive integer):

https://web2.0calc.com/questions/how-many-3s#r21

. i told him that between 0 and 10ⁿ-1 ,every digit (if we write it the way i write the numbers) appears n*10^n-1 times, meaning every digit except for 0 appears n*10^n-1 (but we dont care about the zeroes). what we want to find is the number of times every digit (except for 0) appears between 0 and 999,999=10⁶-1. so, the answer is 6*10^6-1=6*10⁵.

so, P(k)=6*10⁵ for every k that is between 1 and 9.

P(0)*0+P(1)*1+P(2)*2+....+P(9)*9+1=P(1)*1+P(2)*2+....+P(9)*9+1=6*10⁵(1+2+3+4....+9)+1=6*10⁵*45+1=27,000,000+1=

27,000,001

I think i got the answer.

The sequence can be described by the following formula: a_n+2=a_n*(n+2), a₁=1 and a₂=2

That means the next 3 terms are: 384*10=3840, 945*11=10395, 3840*12=46080

But there is something else i want to add about this type of questions:

I feel its a little bit silly, because you could always find another pattern or another way. Of course, if you need to answer that type of questions in a test, you cant write that, you'll have to do what they expect you to do. But i feel its wrong NOT to mention what im going to.

I once heard a joke about a mathematician that has an interesting way of answering those questions: he simply writes "19" when he's asked to find the next term. He justifies that with one of the theorems of a famous mathematician called Lagrange. And in my opinion he's right. Saying that there is a finite amount of answers is silly, because there isnt. There is always a way to make order out of what seems to be chaos.

I dont know if you love math or care about it, but if you do, remember it.