1+2+3+4+5+6+7+8+9+10= 55. Is there a quick formula to calculate those kinds of sums?
Yes. There is a story about Isaac Newton. When he was in school, the teacher asked the pupils to sum up the numbers from 1 to 100. All the pupils began adding up the numbers 1 by 1, but Isaac thought for a minute and then put up his hand. He said "The sum is 5050". Astonished, the teacher asked him how he go the answer so fast. Isaac said, "well I added 1 to 100 and got 101, then I added 2 to 99 and got 101, the I added 3 to 98 and got 101. I figured out there must be 50 such pairs, so I multiplied 101 by 50 and got 5050".
Given that; 1+2+3+4+5+6+7+8+9+10 = 25
The quick formular is to apply Isaac Newton.
1+10=11
2+9=11
3+8=11
4+7=11
5+6=11
(5 pairs)
Therefore 5 multiple by 11 is equal to; 55
So if you want the sum of numbers in sequence starting from 1, if there are n numbers, the sum is
(n + 1) * n/2
For example, 1 to 8:
9 X 4 = 36
For 1 to 9:
10 X 4.5 = 45
for example:
calculating the sequence of 9, would be 1+2+3+4+5+6+7+8+9= 45
For 5 would be 1+2+3+4+5= 15
For 8, it would be= 1+2+3+4+5+6+7+8= 64
Use the above menthol to solve these examples in the comment box. Let see if you can get it right correctly
What others method can you use to solve this problem quickly? Comment below
Comments
Post a Comment