Question attatched. Im not really sure how to approach these types of quesitons. I can do these if theyre only one group or two groups with much lower powers (then id just expand it). Quickest way of doing this type of question?
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Binomials - coeffeciants of terms (1 Viewer)
- Thread starter Mumma
- Start date
For this one it's probably best to combine the brackets a bit to get:
(y<sup>2</sup> - 1/y<sup>2</sup>)<sup>7</sup>(y + 1/y)<sup>3</sup>
= (y<sup>2</sup> - 1/y<sup>2</sup>)<sup>7</sup>(y<sup>3</sup> + 3y + 3/y + y<sup>-3</sup>)
In the left bracket the first power in the expansion is y^14 which then goes down by 4 each time to y^10, y^6 ... y^-2, y^-6 ...
Look for power combinations between the brackets which will give you '-3' and you should see y<sup>-2</sup>.y<sup>-1</sup> = y<sup>-3</sup> = y<sup>-6</sup>.y<sup>3</sup> ... Then you just have to work out the coefficients of the relevant powers, multiply them and add:
coefficient(y<sup>-2</sup>) x coefficient(y<sup>-1</sup>) + coefficient(y<sup>-6</sup>) x coefficient(y<sup>3</sup>)
= <sup>7</sup>C<sub>4</sub>.3 + <sup>7</sup>C<sub>5</sub>.1
= 126
Hopefully that should all be right. Let us know if you need any more explanation.
(y<sup>2</sup> - 1/y<sup>2</sup>)<sup>7</sup>(y + 1/y)<sup>3</sup>
= (y<sup>2</sup> - 1/y<sup>2</sup>)<sup>7</sup>(y<sup>3</sup> + 3y + 3/y + y<sup>-3</sup>)
In the left bracket the first power in the expansion is y^14 which then goes down by 4 each time to y^10, y^6 ... y^-2, y^-6 ...
Look for power combinations between the brackets which will give you '-3' and you should see y<sup>-2</sup>.y<sup>-1</sup> = y<sup>-3</sup> = y<sup>-6</sup>.y<sup>3</sup> ... Then you just have to work out the coefficients of the relevant powers, multiply them and add:
coefficient(y<sup>-2</sup>) x coefficient(y<sup>-1</sup>) + coefficient(y<sup>-6</sup>) x coefficient(y<sup>3</sup>)
= <sup>7</sup>C<sub>4</sub>.3 + <sup>7</sup>C<sub>5</sub>.1
= 126
Hopefully that should all be right. Let us know if you need any more explanation.