How to find the sum ?

Question

How to find the sum ?

1 + 3 – 5 + 7 + 9 – 11 + 13 + 15 – 17 + … + 595 + 597 – 599 ?

Answer 1

Break it down into three sums (A + B + C):
A: 1 + 7 + 13 + 19 + ... + 595
B: 3 + 9 + 15 + 21 + ... + 597
C: -5 - 11 - 17 - 19 - ... - 599

A = sum(i=0 to 99)(6i + 1)
B = A + sum(i=0 to 99)(2)
C = -1 * sum(i=0 to 99)(6i + 5)

A = sum(i=0 to 99)(6i) + sum(i=0 to 99)(1)
B = A + sum(i=0 to 99)(2)
C = -1 * [sum(i=0 to 99)(6i) + sum(i=0 to 99)(5)]

A = 6*sum(i=0 to 99)(i) + 100
B = A + 200
C = -1 * [6*sum(i=0 to 99)(i) + 500]

sum(i=0 to 99)(i) = (99)(100)/2 = 4950

A = 6*4950 + 100
B = 6*4950 + 100 + 200
C = -1*[6*4950 + 500]

A = 29800
B = 30000
C = -30200

A + B + C = 29600

Answer 2

Arithmetic sequence:

Sum(n) = x1 + x2 + ... + xn = n * (x1 + xn) / 2
xn = x1 + (n -1)d
n = (xn - x1)/d +1, (d is delta)

=> Sum(n) = ((xn - x1)/d +1) (x1 + xn) / 2

1 + 3 – 5 + 7 + 9 – 11 + 13 + 15 – 17 + … + 595 + 597 – 599
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + ... + 595 + 597 + 599 - 5 - 11 - 17 - ... - 593 - 599 - 5 - 11 - 17 - ... - 593 - 599
= 1 + 3 + 5 + 7 + ... + 599 - 2 * ( 5 + 11 + 17 + ... + 593 + 599)
= x - 2y

where

x = 1 + 3 + 5 + 7 + ... + 599
= ((599-1)/2+1)(1+599)/2
= 300 * 600 / 2
= 90000

y = 5 + 11 + 17 + ... + 593 + 599
= (599-5)/6+1)(5+599)/2
= 100 * 604 /2
= 30200

So the answer is
1 + 3 – 5 + 7 + 9 – 11 + 13 + 15 – 17 + … + 595 + 597 – 599
= 90000 - 2 * 30200
= 29600

Answer 3

find a patern and explotit it
in 1+3-5= -1
then -1 +7+9-11= 4
then 4+13+15-17=15
still not seeing a patern? then continue
15+19+21-23=32
32+25+27-29=55
at this point I'd graph it to see if there would be a line that'd fit (which there should be)
baring that we look at the differnce between the answers after subtacttion
5 between -1 and 4
11 between 4 and 15
17 betwwen 15 and 32
23 between 32 and 55
and we have a corelation... it'd appear like if we add all of the numbers that would be subtracted , subtract 1 and don't add the last one we'd get the answer
so that'd be 5+11+17+23+29+35+41+47+53+59+65+71+77+83+89+95+101+107... -1
or
we can even breack it down even more if you'll notice the numbers involed are each one less the a power of 6 or in math turms
(6-1)+(2*6-1)+(3*6-1)+...
of wich we can further say (6-1)*99+(6*(97+96+95+...))-1
(5*99)+(6*4698)-1
495+28188-1=28682

Answer 4

599=2·299+1

1+3-5=4-5=-1
7+9-11=16-11=5 --> d1=6
13+15-17=11 --> d2=6
19+21-23=40-23=17 --> d2=6......

--> bn=6n-7

bn= - [ (6n-1)-(6n-3)-(6n-5)] = 6n -7
b1= - 5+3+1=-1
b2= -11+9+7= 5
.....

b100= - 599+597+595=593

--> 1 + 3 – 5 + 7 + 9 – 11 + 13 + 15 – 17 + … + 595 + 597 – 599 = b1+b2+...+b100 = [P.a.] = (b1+b100)·n/2
=(-1+593)·50= 29600

Saludos.

Answer 5

The positive numbers form the atrithmetic sequence:
4,12, 28, 40 ... where d= 12.
The negative numbers form the arithmetic sequence:
-5, -11, -17, ... where d = -6
So just use the standard formula for the sum of an arithmetic series and add the results together.

Answer 6

use steps

1=3+4
4-5=-1
-1+7= 6+9=15
15-11+4
4+13=17+15=32
32-17=25

( you gewt how to do it)

Answer 7

I presume these are all WHOLE NUMBERS...... So, simply add each (recognising each number's polarity (+ / -) beginning from the beginning value and work to the right.

Answer 8

this is very confusing but i think the answer is 13

Answer 9

by adding all the numbers together

Answer 10

29,600