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Re :addition of exponents [#permalink]
28 Apr 2006, 13:43

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

Re: Re :addition of exponents [#permalink]
28 Apr 2006, 15:25

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...

Re: Re :addition of exponents [#permalink]
28 Apr 2006, 15:27

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?

Re: Re :addition of exponents [#permalink]
28 Apr 2006, 20:08

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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