Strange Sum
1 + 2 − 3 − 4 + 5 + 6 − 7 − 8 + … + 3 0 1 + 3 0 2 = ?
Clarification : The sum keeps alternating between two distinct positive numbers and two distinct negative numbers.
The answer is 303.
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8 solutions
Nice solution! Can we generalize this summation up to n terms?
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H'mmm. For any number of terms that's an exact multiple of 4, say 4n terms, the answer is clearly -4n. Since the next term is (4n+1) added, for 4n+1 terms the answer is 1. the next term is (+ 4n+2) so that yields, for 4n+2 terms,an answer of 4n+3. And the next term is subtracted (4n+3), so for any number (4n+3) terms the answer is 0. that covers every case.
( - 0 + 1 +2 - 3 ) = 0 , ( - 4 + 5 + 6 - 7 ) = 0 , .................................. ( - 296 + 297 + 298 - 299 ) = 0 that's 75 grouping it's sum = 0.0 ( - 300 + 301 + 302 ) = 303 OK
Nice solution !
Can't understand how 302 came
As I find that the mean of (a)+(a+1) equals to (a-1)+(a+2) , hence 2-3-4+5 equals to 0, and 6-7-8+9 equals to zero, and vice versa, hence the answer is 1+302, which is 303
Nice solution!
(1 + 2 - 3) = 0
(- 4 + 5 + 6 - 7) = 0 (- 8 + 9 + 10 - 11) = 0 ... (- 300 + 301 + 302) = 303
or
(- 4 + 5 ) = 1 < = > (- 8 + 9 ) = 1 < = > ... < = > (-300 + 301) = 1
1 + 302 = 303
1+(2-3-4+5)=1+0 (6-7-8+9)+(10-11-12+13)... +(298-299-300+301)=0
So
1+302 = 303
You can gather every 4 terms into 1 new terms, i.e.: ( 1 + 2 − 3 − 4 ) , ( 5 + 6 − 7 − 8 ) , . . .
then you will get bunch(s) of − 4 . To count the the number of terms (of -4), you will find the greatest integer (close to 302) that's divisible by 4, that is 300. So, there are 4 3 0 0 = 7 5 terms of − 4 then you will have 3 0 1 + 3 0 2 as the "left-over".
So, the total sum is: 7 5 ∗ ( − 4 ) + 3 0 1 + 3 0 2 = 3 0 3
1 + 2 − 3 − 4 + 5 + 6 − 7 − 8 + … + 3 0 1 + 3 0 2
1 + 3 0 2 = 3 0 3
2 + 3 0 1 = 3 0 3
− 3 − 3 0 0 = − 3 0 3
− 4 − 2 9 9 = − 3 0 3
The sum of these numbers cycle up to 606 back to 0 every 4 pairs
Total number of pairs = 2 1 3 0 2 = 1 5 1
1 5 1 ( m o d 4 ) = 3
This means we just need the sum of the first 3 pairs
3 0 3 + 3 0 3 − 3 0 3 = 3 0 3
Before the term 5 , the sum of the first 4 terms is − 4 . Before the term 9 , the sum of the first 8 terms is − 8 . Therefore, the sum of all the terms before the term 3 0 1 is − 3 0 0 .
S = − 3 0 0 + 3 0 1 + 3 0 2 = 3 0 3
( 1 + 2 − 3 − 4 ) = − 4 , ( 5 + 6 − 7 − 8 ) = − 4 , … , ( 2 9 7 + 2 9 8 − 2 9 9 − 3 0 0 ) = − 4 .
That's 75 groupings of -4. that's a total of -300.
(-300) + 301 = 1
1 + 302 = 303