Find the value of 1 + 2 + 3 - 4 + 5 + 6 + 7 - 8 + ... + 197 + 198 + 199 - 200.
+1005 Patterns.
1+2+3-4 = 2
5+6+7-8 = 10
9+10+11-12 =18
197+198+199-200 = 394
The pattern goes like this
2,10,18... 394
The new num. is increased by 8 every iteration. Each "interation" is 4 numbers, 200/4 = 50.
You have a pattern with 50 digits, starting with 2, and each new digit is 8 more than the previous.
See if you can do this.
If you don't understand anything feel free to ask
(6 - 4) + (18 - 8) + (30 - 12) + (42 - 16)+............+ (594 - 200) = sum_(n=1)^50(8 n - 6) = 9900
+1005 Yeah, also it's possible to do this:
2+10+18...+394
Yup it's 9900
I was too lazy to do the math so i just wrote a java program lol
int x = 2;
int y = 0;
for(int i = 0; i < 50; i++) {
y+=x;
x+=8;
println(y);
}
+128460 Thanks , hugo.....here's another way
The sum of the first 200 positive integers is just (200)(201) / 2 = 100 * 201 = 20100
But we have actually "over-summed" the given series by
2 [ 4 + 8 + 12 + 16 +.....+ 200 ]
The number of terms in the brackets = [ 200 - 4 ] / 4 + 1 = 196/4 + 1 = 50
So the sum of this series is just
2 [ 200 + 4 ] * (50/2] = 204 * 50 = 10200
So.....the sum of this series = 20100 - 10200 = 9900
+1005 Smart. I like it.
[inserts appropriate GIF]
Yours is much better than mine
hehehe
