45
The sum of all natural numbers 0+2+2+3+4+5+6+7+8+9... = -1/12
The answer is: -2, -1, 0, 1 as (-2)+(-1)+0+1 = -2
The numbers are -2, -1, 0, 1 and 2.
its 5035the summarian notation tells you that(sum of all #from 0 to a number'N'(sum of all #from 0 to N) = (n)+(n-1)+(n-2)+(n-3)+...+(2) +(1) or(sum of all #from 0 to N) = (1)+ (2) + (3) + (4)+...+(n-1)+(n)the two different sums are aligned by columns. now add the two colunms accordingly and you'll get2x(sum of all #from 0 to N)=(n+1)+(n+1)+...+(n+1) (n+1)is added n timesso2x(sum of all #from 0 to N) =n(n+1)(sum of all #from 0 to N) =n(n+1)/2so (sum of 5 to 100) = (sum of 0 to 100)- (sum of 0 to 5)=100(101)/2 - 5(6)/2 = 5050 - 15 = 5035
-2, -1, 0, 1
Do this in reverse. The sum of -1 and -1 is -1+-1=-2 The difference of -6 and -6 is -6-(-6)=-6+6=0 0 increased by -2 is 0+-2=-2 The sum of 10 and -2 is 10+-2=8 Translation: 8
2+(-2) =0
The sum of all natural numbers 0+2+2+3+4+5+6+7+8+9... = -1/12
The answer is: -2, -1, 0, 1 as (-2)+(-1)+0+1 = -2
The numbers are -2, -1, 0, 1 and 2.
:−3+0+1+2 = 0 :−3*0*1*2 = 0
its 5035the summarian notation tells you that(sum of all #from 0 to a number'N'(sum of all #from 0 to N) = (n)+(n-1)+(n-2)+(n-3)+...+(2) +(1) or(sum of all #from 0 to N) = (1)+ (2) + (3) + (4)+...+(n-1)+(n)the two different sums are aligned by columns. now add the two colunms accordingly and you'll get2x(sum of all #from 0 to N)=(n+1)+(n+1)+...+(n+1) (n+1)is added n timesso2x(sum of all #from 0 to N) =n(n+1)(sum of all #from 0 to N) =n(n+1)/2so (sum of 5 to 100) = (sum of 0 to 100)- (sum of 0 to 5)=100(101)/2 - 5(6)/2 = 5050 - 15 = 5035
Well, well, well, look who's trying to do some math! If the sum of ten consecutive integers is five, then the middle number is 0. So, the largest integer would be 4. Keep crunching those numbers, honey!
4*4 = 16 (4+4 = 8)3*5 = 15 (3+5 = 8)2*6 = 12 (2+6 = 8)1*7 = 7 (1+7 = 8)0*8 = 0 (0+8 = 8)So 4 and 4 produce the largest product and still have the sum of 8.Assumptions:only integers,that this can be extrapolated (-1*9 < 0*8 etc)
-2, -1, 0, 1
2,2,0,5,1,4,1,3,0,0,1,4,4,0,1,4,3,4,2,1,0
Let the five consecutive integers be ( x-2, x-1, x, x+1, x+2 ). The sum of these integers can be expressed as ( (x-2) + (x-1) + x + (x+1) + (x+2) = 5x ). To find the integers that sum to 0, we set ( 5x = 0 ) which gives ( x = 0 ). Thus, the five consecutive integers are -2, -1, 0, 1, and 2.