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If x=7, what is the value of 1-x^2+x^3-x^4+x^5-x^6

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Director
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If x=7, what is the value of 1-x^2+x^3-x^4+x^5-x^6 [#permalink]

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28 Apr 2006, 03:08
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If x=7, what is the value of 1-x^2+x^3-x^4+x^5-x^6?

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28 Apr 2006, 08:05
joemama142000 wrote:
If x=7, what is the value of 1-x^2+x^3-x^4+x^5-x^6?

Isnt it true that we cant simply exp when adding or subtracting? Confused.

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28 Apr 2006, 08:11
shampoo wrote:
joemama142000 wrote:
If x=7, what is the value of 1-x^2+x^3-x^4+x^5-x^6?

Isnt it true that we cant simply exp when adding or subtracting? Confused.

oh stupid me, since we know x=7 we can jus sub-in and solve. any easier way to do this without counting?

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Manager
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28 Apr 2006, 08:38
i simplified first to:
6(8 + 7^3 + 7^5)

got 102,948

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Intern
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28 Apr 2006, 14:43
The question asks for sum of
1-x^2+X^3-X^4+x^5-X^6
In effect it is a geometric progression after 1 each term being multiplied by -X
so 1-(X^2-X^3+X^4-X^5+X^6) =1- Sum of geometric progression

Sum of geometric progression =a(1-r^n)/1-r

Where a= the first term which in this case is X^2
And r = the common ratio which in this case is â€“X
And n=the number of terms which in this case is 5 (X^2-------X^6)
So substituting we get
S=X^2 (1+X^5) /1+X in other words 49x 16808/8

=49x2101=102949
1-S=1-102949=-102948

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VP
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28 Apr 2006, 16:25
fighter wrote:
The question asks for sum of
1-x^2+X^3-X^4+x^5-X^6
In effect it is a geometric progression after 1 each term being multiplied by -X
so 1-(X^2-X^3+X^4-X^5+X^6) =1- Sum of geometric progression

Sum of geometric progression =a(1-r^n)/1-r

Where a= the first term which in this case is X^2
And r = the common ratio which in this case is â€“X
And n=the number of terms which in this case is 5 (X^2-------X^6)
So substituting we get
S=X^2 (1+X^5) /1+X in other words 49x 16808/8

=49x2101=102949
1-S=1-102949=-102948

Hmm, I've never seen this formula before. Great job tho Can anybody enlighten us with this formula?
_________________

Don't be afraid to take a flying leap of faith.. If you risk nothing, than you gain nothing...

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Manager
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28 Apr 2006, 16:27
TeHCM wrote:
fighter wrote:
The question asks for sum of
1-x^2+X^3-X^4+x^5-X^6
In effect it is a geometric progression after 1 each term being multiplied by -X
so 1-(X^2-X^3+X^4-X^5+X^6) =1- Sum of geometric progression

Sum of geometric progression =a(1-r^n)/1-r

Where a= the first term which in this case is X^2
And r = the common ratio which in this case is â€“X
And n=the number of terms which in this case is 5 (X^2-------X^6)
So substituting we get
S=X^2 (1+X^5) /1+X in other words 49x 16808/8

=49x2101=102949
1-S=1-102949=-102948

Hmm, I've never seen this formula before. Great job tho Can anybody enlighten us with this formula?

check this link, it explains it very well:

http://mathworld.wolfram.com/GeometricSeries.html

Last edited by conocieur on 28 Apr 2006, 16:28, edited 1 time in total.

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Senior Manager
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28 Apr 2006, 21:08
fighter wrote:
The question asks for sum of
1-x^2+X^3-X^4+x^5-X^6
In effect it is a geometric progression after 1 each term being multiplied by -X
so 1-(X^2-X^3+X^4-X^5+X^6) =1- Sum of geometric progression

Sum of geometric progression =a(1-r^n)/1-r

Where a= the first term which in this case is X^2
And r = the common ratio which in this case is â€“X
And n=the number of terms which in this case is 5 (X^2-------X^6)
So substituting we get
S=X^2 (1+X^5) /1+X in other words 49x 16808/8

=49x2101=102949
1-S=1-102949=-102948

Good approach,

I was looking for short cut but ended up calculating 7*6 good things is you only have to multiply 6 times!!! and I had to redo it only twice...

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Director
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29 Apr 2006, 14:57
figher , that is a great explanation

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Manager
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30 Apr 2006, 01:07
kook44 wrote:
i simplified first to:
6(8 + 7^3 + 7^5)

got 102,948

Explain how you simplified to this

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Manager
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30 Apr 2006, 03:51
Fighter,

What happens if this is an infinite series, we don't know the value of N.

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30 Apr 2006, 03:51
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If x=7, what is the value of 1-x^2+x^3-x^4+x^5-x^6

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