Question 934953
Rearranging that sum so that each numerator is greater that the one before, I get
{{{4/n^4+5/n^4+6/n^4+"..."+(n^4-6)/n^4+(n^4-5)/n^4+(n^4-4)/n^4=309}}}
{{{(4+5+6+"..."+(n^4-6)+(n^4-5)+(n^4-4))/n^4=309}}}
The numerator in the expression above is the sum of {{{n^4-7}}} terms of an arithmetic sequence with common difference {{{1}}} ,
first term {{{4}}} , and last term {{{n^4-4}}} .
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
{{{((4+n^4-4)/2)*(n^4-7)=(n^4/2)*(n^4-7)=n^4(n^4-7)/2}}} ,
so we can re-write the equation as
{{{(n^4(n^4-7)/2)/n^4=309}}} , which simplifies to
{{{(n^4-7)/2=309}}}
Solving for a positive integer {{{n}}} ,
{{{(n^4-7)/2=309}}}--->{{{n^4-7=2*309}}}--->{{{n^4-7=618}}}--->{{{n^4=7+618}}}--->{{{n^4=625}}}--->{{{highlight(n=5)}}} .
If {{{n}}} did not need to be positive, {{{n=-5}}} would also be a solution.