Binomial Theorem (1 Viewer)

allGenreGamer

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I have a few questions in the binomial theorem topic which I cannot solve:

- Write down the (k + 2)th term of (a + b)^2n
- Find the coefficient of a^-2 in the expansion of (a+1/a)^3 * (a - 1/a) ^ 5

These questions are from Fitzpatrick - the one who writes the hardest maths problems! :) I feel that if I can solve these questions, then the coming exam will be no problem, please help me.
 

CM_Tutor

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First Question:

For the expansion (a + b)<sup>2n</sup>, the (k + 1)th term, T<sub>k+1</sub> is T<sub>k+1</sub> = <sup>2n</sup>C<sub>k</sub>a<sup>2n-k</sup>b<sup>k</sup>
Replacing (k + 1) with (k + 2), T<sub>k+2</sub> = <sup>2n</sup>C<sub>k+1</sub>a<sup>2n-(k+1)</sup>b<sup>k+1</sup> = <sup>2n</sup>C<sub>k+1</sub>a<sup>2n-k-1</sup>b<sup>k+1</sup>
 

CM_Tutor

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Second Question:

[a + (1 / a)]<sup>3</sup> = <sup>3</sup>C<sub>0</sub>a<sup>3</sup> + <sup>3</sup>C<sub>1</sub>a<sup>2</sup>(1 / a)<sup>1</sup> + <sup>3</sup>C<sub>2</sub>a<sup>1</sup>(1 / a)<sup>2</sup> + <sup>3</sup>C<sub>3</sub>(1 / a)<sup>3</sup> = <sup>3</sup>C<sub>0</sub>a<sup>3</sup> + <sup>3</sup>C<sub>1</sub>a + <sup>3</sup>C<sub>2</sub> / a + <sup>3</sup>C<sub>3</sub> / a<sup>3</sup> ___ (A)

[a - (1 / a)]<sup>5</sup> = <sup>5</sup>C<sub>0</sub>a<sup>5</sup> + <sup>5</sup>C<sub>1</sub>a<sup>4</sup>(-1 / a)<sup>1</sup> + <sup>5</sup>C<sub>2</sub>a<sup>3</sup>(-1 / a)<sup>2</sup> + ... + <sup>5</sup>C<sub>5</sub>(-1 / a)<sup>5</sup> = <sup>5</sup>C<sub>0</sub>a<sup>5</sup> - <sup>5</sup>C<sub>1</sub>a<sup>3</sup> + <sup>5</sup>C<sub>2</sub>a - ... - <sup>5</sup>C<sub>5</sub> / a<sup>5</sup> ___ (B)

Every term in the expansion of [a + (1 / a)]<sup>3</sup>[a - (1 / a)]<sup>5</sup> is formed by multiplying a term in (A) by a term in (B). We seek the term in a<sup>-2</sup>. This term is:

Term in a<sup>3</sup> in (A) * Term in a<sup>-5</sup> in (B) + Term in a in (A) * Term in a<sup>-3</sup> in (B) + Term in a<sup>-1</sup> in (A) * Term in a<sup>-1</sup> in (B) + Term in a<sup>-3</sup> in (A) * Term in a in (B)
= [<sup>3</sup>C<sub>0</sub>a<sup>3</sup> * (-<sup>5</sup>C<sub>5</sub> / a<sup>5</sup>)] + [<sup>3</sup>C<sub>1</sub>a * (<sup>5</sup>C<sub>4</sub> / a<sup>3</sup>)] + [(<sup>3</sup>C<sub>2</sub> / a) * (-<sup>5</sup>C<sub>3</sub> / a)] + [(<sup>3</sup>C<sub>3</sub> / a<sup>3</sup>) * <sup>5</sup>C<sub>2</sub>a]
= a<sup>-2</sup>(-<sup>3</sup>C<sub>0</sub> * <sup>5</sup>C<sub>5</sub> + <sup>3</sup>C<sub>1</sub> * <sup>5</sup>C<sub>4</sub> - <sup>3</sup>C<sub>2</sub> * <sup>5</sup>C<sub>3</sub> + <sup>3</sup>C<sub>3</sub> * <sup>5</sup>C<sub>2</sub>)
= a<sup>-2</sup>(-1 * 1 + 3 * 5 - 3 * 10 + 1 * 10)
= -6a<sup>-2</sup>

So, the coefficient of a<sup>-2</sup> in the expansion of [a + (1 / a)]<sup>3</sup>[a - (1 / a)]<sup>5</sup> is -6.
 

allGenreGamer

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Thanks! With that explanation, I've got my head around the problem. Alrite... getting confident for the test! :p
 

masteraal

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how retarded does it look on the screen. very hard to explain without writing it on paper eh :p
 

Xayma

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Originally posted by masteraal
how retarded does it look on the screen. very hard to explain without writing it on paper eh :p
Your just lucky he uses html tags. Imagine it without them ;)
 

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