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# help with sum

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Find the value of 1 + 2 + 3 - 4 + 5 + 6 + 7 - 8 + ... + 197 + 198 + 199 - 200.

May 5, 2020

#1
-4

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

May 5, 2020
#2
+1

(6 - 4) + (18 - 8) + (30 - 12) + (42 - 16)+............+ (594 - 200) = sum_(n=1)^50(8 n - 6) = 9900

May 5, 2020
#4
-4

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);
}

hugomimihu  May 5, 2020
#3
+2

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   May 5, 2020
edited by CPhill  May 5, 2020
#5
-4

Smart. I like it.

[inserts appropriate GIF]

Yours is much better than mine

hehehe

hugomimihu  May 5, 2020