SOLUTION: The sum of the first n integers 1 + 2 + 3 + ... + n = n(n + 1)/2 a How many numbers must be taken to have a sum greater than one million? b Why can’t the sum ever equal 10

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Question 1086342: The sum of the first n integers 1 + 2 + 3 + ... + n =
n(n + 1)/2 a How many numbers must be taken to have a sum greater than one million? b Why can’t the sum ever equal 100 000?
Found 2 solutions by Fombitz, Edwin McCravy:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
%28n%28n%2B1%29%29%2F2%3E1000000
n%28n%2B1%29%3E2000000
n%5E2%2Bn%3E2000000
n%5E2%2Bn-2000000%3E0
We can solve for when it equals zero and then take the next largest integer,
n%5E2%2Bn-2000000=0
%28n%5E2%2Bn%2B1%2F4%29=2000000%2B1%2F4
%28n%2B1%2F2%29%5E2=8000001%2F4
n%2B1%2F2=0+%2B-+sqrt%288000001%29%2F2
n=1%2F2+%2B-+%283sqrt%28888889%29%29%2F2
Only positive n makes sense in this problem,
n=%281%2B3sqrt%28888889%29%29%2F2
or approximately,
n=1413.71
So the next largest integer is,
N=1414
.
.
.
Assume the sum does equal 100000,
%28n%28n%2B1%29%29%2F2=10000
n%5E2%2Bn=200000
n%5E2%2Bn-200000=0
Similarly,
%28n%2B1%2F2%29%5E2=200000%2B1%2F4
%28n%2B1%2F2%29%5E2=800001%2F4
Since the sqrt%28800001%29 is not an integer, there is no hope that the required sum would be an integer.
So it would never equal 100000.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
%28n%28n%2B1%29%29%2F2%22%22%3E%22%22}1000000

Multiply through by 2:

n%28n%2B1%29%22%22%3E%22%22}2000000

n%5E2%2Bn%29%22%22%3E%22%22}2000000


n%5E2%2Bn-2000000%29%22%22%3E%22%22}0

Use the quadratic formula to find the critical numbers

%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

%28-1+%2B-+sqrt%28+1%5E2-4%2A1%2A%28-2000000%29+%29%29%2F%282%2A1%29+

%28-1+%2B-+sqrt%28+1%2B8000000%29+%29%29%2F2+

%28-1+%2B-+sqrt%288000001%29+%29%29%2F2+

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%28n%2B1%29%2F2

1413%281413%2B1%29%2F2=998991, 1414%281414%2B1%29%2F2=1000405

-----------------

Why cant the sum ever equal 100000.

Suppose for contradiction, that it can = 100000

%28n%28n%2B1%29%29%2F2%22%22=%22%22}100000

Multiply through by 2:

n%28n%2B1%29%22%22=%22%22}200000

n%5E2%2Bn%29%22%22=%22%22}200000

n%5E2%2Bn-200000%29%22%22=%22%22}0

Use the quadratic formula to solve for n

n=%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

n=%28-1+%2B-+sqrt%28+1%5E2-4%2A1%2A%28-200000%29+%29%29%2F%282%2A1%29+

n=%28-1+%2B-+sqrt%28+1%2B800000%29+%29%29%2F2+

n=%28-1+%2B-+sqrt%28800001%29+%29%29%2F2+

n=%28-1+%2B-+sqrt%28800001%29+%29%29%2F2+

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%28n%2B1%29%2F2

446%28446%2B1%29%2F2=99681, 447%28447%2B1%29%2F2=100128.

So the sum can never be 100000.

Edwin