- #1

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I went [tex]t_{k+1}= _{14}C_{k}(2x)^{14-k}(-x^{4})^{k}[/tex]

[tex]= 2x^{14-k}(-x^{4k})[/tex]

First of all, am I on the right track? If so what exactly do I do from there?

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- Thread starter ms. confused
- Start date

- #1

- 91

- 0

I went [tex]t_{k+1}= _{14}C_{k}(2x)^{14-k}(-x^{4})^{k}[/tex]

[tex]= 2x^{14-k}(-x^{4k})[/tex]

First of all, am I on the right track? If so what exactly do I do from there?

- #2

AKG

Science Advisor

Homework Helper

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[tex]t_{k+1} = {{14}\choose k}(-2)^{14-k}x^{14 + 3k}[/tex]

See if you can figure out why the above is right. Now, to find the term containing x

(91)(2

- #3

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- #4

lurflurf

Homework Helper

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You want the x^20 term so if the exponent of x in the kth term isms. confused said:

4k+(14-k)=20 what is k?

- #5

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+3k... right...thanks I see it now!

- #6

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Take (x^14) common and find the third term in the expansion of (2-(x^3))^14 i.e.

14C2*(2^12)*(-1)^2.

14C2*(2^12)*(-1)^2.

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