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?

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) (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

The numerator in the expression above is the sum of terms of an arithmetic sequence with common difference ,
first term , and last term .
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
, which simplifies to

Solving for a positive integer ,
--->--->--->--->---> .
If did not need to be positive, would also be a solution.
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