anonymous
  • anonymous
QUESTION FOR FUN SOLVE FOR n: n^n +n^(n-1) + n^(n-2)+...+n^2+n^1+n^0 = n^n^n
Mathematics
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

shubhamsrg
  • shubhamsrg
1 + n + n^2 .. n^n = n^n^n n=1 is a clear solution for n not equal to 1, we have => (n^(n+1) -1)/(n-1) = n^n^n unsure what will come next.
anonymous
  • anonymous
1^1 +1^0 = 1^1^1 2=1 which is FALSE
anonymous
  • anonymous
So, n=1 is not a solution

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Should I post my solution?
shubhamsrg
  • shubhamsrg
Oh,,Sorry about that.
shubhamsrg
  • shubhamsrg
Well I failed to do it doesn't mean no one will do it. Many geniuses are there on OS. I'd recommend you to wait. :)
anonymous
  • anonymous
As u say
anonymous
  • anonymous
2 minutes, I'm having a crack.
anonymous
  • anonymous
U can take your time
anonymous
  • anonymous
Before I spend the next hour looking at it to find I cant find a solution, is there definately solutions?
anonymous
  • anonymous
Apologise if it is obvious to have solutions, its been a long day of uni
anonymous
  • anonymous
Actually, NO SOLUTION
anonymous
  • anonymous
But u need to PROVE it
anonymous
  • anonymous
We sure? Do you have a explanation/proof to that?
anonymous
  • anonymous
Ahh okay
anonymous
  • anonymous
I guess I have got one proof...
shubhamsrg
  • shubhamsrg
BY no solution, you mean no integral solution right ?
anonymous
  • anonymous
The proof that I have got is very simple
anonymous
  • anonymous
Shoot.
anonymous
  • anonymous
I will message u if u want
anonymous
  • anonymous
I'd like to see it. I feel there is a solution as n^n^n grows faster than a quadratic and the quadratic is smaller than the exponent around n=0
shubhamsrg
  • shubhamsrg
WOlfram tells me a solution, You must be meaning no integral solutions are there hmm ?
anonymous
  • anonymous
How did you enter that into wolfram? I couldnt get it to work.
shubhamsrg
  • shubhamsrg
http://www.wolframalpha.com/input/?i=%28n%5E%28n%2B1%29+-1%29%2F%28n-1%29+%3D+n%5En%5En
anonymous
  • anonymous
Sorry, I mean integer solution of n
shubhamsrg
  • shubhamsrg
hmm, that is what i assumed.
anonymous
  • anonymous
Well from the question n^n +n^(n-1) + n^(n-2)+...+n^2+n^1+n^0 = n^n^n it is clear n>=0 and n is integer
anonymous
  • anonymous
So yea we agreed there is no solution? as the only intersection is irrational.
anonymous
  • anonymous
Now, u just need to prove it
anonymous
  • anonymous
Surely this would suffice: n^n +n^(n-1) + n^(n-2)+...+n^2+n^1+n^0>n^n^n \[\forall n \le 1\] and n^n +n^(n-1) + n^(n-2)+...+n^2+n^1+n^0
anonymous
  • anonymous
ge is supposed to read as >=
anonymous
  • anonymous
But u directly said it without any proof
anonymous
  • anonymous
Let A=n^n +n^(n-1) + n^(n-2)+...+n^2+n^1+n^0, and let B=n^n^n. Note: n is belonging arbitrarily to the set of integers. Then since A1, i.e. all n>=2, and A>B for all n<=1 then since there exists no integer between 1 and 2, there is no possible integer solutions to A=B
anonymous
  • anonymous
Since R.H.S is divisible by n, L.H.S must also be divisible by n. Now, all the terms except n^0 of L.H.S is divisible by n for all values of n. So,L.H.S is divisible by n only if n^0 is also divisible by n. n^0 is divisible by n only when n=1 but n=1 is not the solution. THUS, NO SOLUTION
anonymous
  • anonymous
Interesting how you look at it mroe from the abstract whilst i can see it only from the geometric side of things.
shubhamsrg
  • shubhamsrg
neatly done @sauravshakya

Looking for something else?

Not the answer you are looking for? Search for more explanations.