Question

evaluate the value of A = ?

A = 1^3 - 2^3 + 3^3 - 4^3 + 5^3 - 6^3 + ... + 99^3

Answer 1

Okay well there is an easy way to do this problem and there is also a hard way. Let me look in my Calculus notes to help you out with this problem.

Answer 2

Okay here we go. Your problem comes down to Sigma E = i^3 where i =1 and n = 99.
You can figure this out with a formula, which i will give you in a second.

Answer 3

still confused what to do next ?

Answer 4

It's n^2(n+1)^2/4

Answer 5

This is your formula you need to use, where n = 99

Answer 6

http://4.bp.blogspot.com/_8-M4eJAFB-U/R_kWJ1h57JI/AAAAAAAAABE/AFFadS130m0/s1600-h/summation+rules.GIF

Answer 7

99^2(100)^2/4 = your answer

Answer 8

Does that make sense?

Answer 9

how do i know it would be equals n^2(n+1)^2/4 ?

Answer 10

This solution is cheating in a way. It is like magic... Formulas appearing from thin air

Answer 11

Because the exponent is 3. If you look at the picture I gave the link to, the other formulas are for if the exponent is 2, 1, or even a constant.

Answer 12

The formula is defined in Summation rules for Sigma Notation

Answer 13

the 4th formula is when all series positive, while the question above is alternately +-+-+-+-...+-+

Answer 14

Ah good point. Well if you will notice the pattern that 1^3 - 99^3 is equivalent 2^3 + 98^3. You only have to divide your answer that you get from the equation in half.

Answer 15

So instead of n^2(n+1)^2/4, you will have n^2(n+1)^2/8

Answer 16

Sorry by equivalent i mean they canel each other out . . . wait a second is the answer 0? What do you think.

Answer 17

in key answer book = 492500 (but didnt show steps)

well, i checked by using your formula :
n^2(n+1)^2/8
99^2 (100)^2 / 8
= 98010000/8
= 12251250

where the mistake ?

Answer 18

I need help. Anyone out there see the problem?

Answer 19

nobody seems to be acknowledging the alternating minus sign

Answer 20

\[\sum_{n=1}^{99}(-1)^{n-1}n^3\]

Answer 21

what's next, @TuringTest ?
that still didnt make sense to me, but thanks trying helps me

Answer 22

Well first, do you understand why I rewrote that series this way?
do you see how the factor of (-1)^(n-1) will get the alternating +/- you have in your series?

Answer 23

just alternative :
it can rewitten as :

A = 1^3 - 2^3 + 3^3 - 4^3 + 5^3 - 6^3 + ... + 99^3

A = 1^3+2^3+3^3+4^3+5^3+6^3+ ...+98^3+99^3 -2(2^3+4^3+6^3+...+98^3)

A = 1^3+2^3+3^3+4^3+5^3+6^3+ ...+98^3+99^3 -2*2^3(1^3+2^3+3^3+...+49^3)

A = 1^3+2^3+3^3+4^3+5^3+6^3+ ...+98^3+99^3 -16(1^3+2^3+3^3+...+49^3)

A = (1+2+3+4+...+99)^2 - 16(1+2+3+4+...+49)^2
A = .....

continue it ...

Answer 24

use the formula what @KirbyLegs said to calculating elements in paratheses
1+2+3+4+...+n = n(n+1)/2

Answer 25

take the sum cube of odd numbers and subtract the sum of cubes of even number. That is,
A = sum{1->50} (2k-1)^3 - sum{1->49} (2k)^3

All you got to do is to expand and use the formula of sum of k, k^2, k^3. The answer is 492,500.
http://www.wolframalpha.com/input/?i=%28sum+k+%3D+1+to+50+%282k-1%29%5E3%29+-+%28sum+k+%3D+1+to+49+%282k%29%5E3%29

Answer 26

A = (1+2+3+4+...+99)^2 - 16(1+2+3+4+...+49)^2
A = (99 * 100/2)^2 - 16(49 * 50/2)^2
A = 4950^2 - 16(1225)^2
A = 24502500 - 24010000
A = 492,500

waw, that's correct answer. Now make sense to me.
Thank you very much every one :D

Answer 27

you're welcome :*