Question 1086342
<pre>
{{{(n(n+1))/2}}}{{{"">""}}}}{{{1000000}}}

Multiply through by 2:

{{{n(n+1)}}}{{{"">""}}}}{{{2000000}}}

{{{n^2+n)}}}{{{"">""}}}}{{{2000000}}}


{{{n^2+n-2000000)}}}{{{"">""}}}}{{{0}}}

Use the quadratic formula to find the critical numbers

{{{(-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{(-1 +- sqrt( 1^2-4*1*(-2000000) ))/(2*1) }}}

{{{(-1 +- sqrt( 1+8000000) ))/2 }}}

{{{(-1 +- sqrt(8000001) ))/2 }}}

That is approximately 1413.713651, between integers 1413 and 1414

So the sum of the first 1413 positive integers is less than 1000000,
and the sum of the first 1414 integers integers is more than 1000000.

To show this we substitute 1413 and 1414 in {{{n(n+1)/2}}}

{{{1413(1413+1)/2=998991}}}, {{{1414(1414+1)/2=1000405}}}

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Why cant the sum ever equal 100000.

Suppose for contradiction, that it can = 100000

{{{(n(n+1))/2}}}{{{""=""}}}}{{{100000}}}

Multiply through by 2:

{{{n(n+1)}}}{{{""=""}}}}{{{200000}}}

{{{n^2+n)}}}{{{""=""}}}}{{{200000}}}

{{{n^2+n-200000)}}}{{{""=""}}}}{{{0}}}

Use the quadratic formula to solve for n

{{{n=(-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{n=(-1 +- sqrt( 1^2-4*1*(-200000) ))/(2*1) }}}

{{{n=(-1 +- sqrt( 1+800000) ))/2 }}}

{{{n=(-1 +- sqrt(800001) ))/2 }}}

{{{n=(-1 +- sqrt(800001) ))/2 }}}

{{{n=446.713875}}}

That's not a positive integer, so that contradicts
our assumption that it could equal to a positive
integer.  So the sum of the first 446 integers is
less than 100000 and the sum of the first 447
integers is more than 100000.

To show this we substitute 446 and 447 in {{{n(n+1)/2}}}

{{{446(446+1)/2=99681}}}, {{{447(447+1)/2=100128}}}.

So the sum can never be 100000.

Edwin</pre>