SOLUTION: 4/n^4+5/n^4+6/n^4+...+(n^4-4)/n^4+(n^4-5)/n^4+(n^4-6)/n^4=309?

Algebra ->  Rational-functions -> SOLUTION: 4/n^4+5/n^4+6/n^4+...+(n^4-4)/n^4+(n^4-5)/n^4+(n^4-6)/n^4=309?      Log On


   

Question 934953: 4/n^4+5/n^4+6/n^4+...+(n^4-4)/n^4+(n^4-5)/n^4+(n^4-6)/n^4=309?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Rearranging that sum so that each numerator is greater that the one before, I get

%284%2B5%2B6%2B%22...%22%2B%28n%5E4-6%29%2B%28n%5E4-5%29%2B%28n%5E4-4%29%29%2Fn%5E4=309
The numerator in the expression above is the sum of n%5E4-7 terms of an arithmetic sequence with common difference 1 ,
first term 4 , and last term n%5E4-4 .
The sum of a number of terms of an arithmetic sequence is
the average of first and last terms
times the number of terms.
In this case, the sum is
,
so we can re-write the equation as
%28n%5E4%28n%5E4-7%29%2F2%29%2Fn%5E4=309 , which simplifies to
%28n%5E4-7%29%2F2=309
Solving for a positive integer n ,
%28n%5E4-7%29%2F2=309--->n%5E4-7=2%2A309--->n%5E4-7=618--->n%5E4=7%2B618--->n%5E4=625--->highlight%28n=5%29 .
If n did not need to be positive, n=-5 would also be a solution.