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# Binomial expansion help please watch

1. Given that the expansion (1+ax)^n begins with 1+36x+576x^2 find the values of a and n
2. (Original post by osayuki)
Given that the expansion (1+ax)^n begins with 1+36x+576x^2 find the values of a and n
Can you expand in terms of and ?

It starts

Then compare the coefficients.

E.g. the second term in the expansion is so .

Does that make sense?
3. (Original post by notnek)
Can you expand in terms of and ?

It starts

Then compare the coefficients.

E.g. the second term in the expansion is so .

Does that make sense?
Thank you for replying. I'm unsure how to get nax as I couldn't figure out how to cancel n!/(n-1!)1!
4. (Original post by osayuki)
Thank you for replying. I'm unsure how to get nax as I couldn't figure out how to cancel n!/(n-1!)1!
(We can "ignore" the 1! since it just equals 1)

Remembering what factorial actually means and writing n! and (n-1)! out fully we get;

Cancelling terms which occur on both lines are left with;

5. (Original post by DylanJ42)
(We can "ignore" the 1! since it just equals 1)

Remembering what factorial actually means and writing n! and (n-1)! out fully we get;

Cancelling terms which occur on both lines are left with;

Thank you so much
6. (Original post by osayuki)
Thank you for replying. I'm unsure how to get nax as I couldn't figure out how to cancel n!/(n-1!)1!
It will make your life easier if you remember the pattern of the coefficients so you don't have to derive each one.

E.g. for the simple expansion (1+x)^n, the second term coefficient is 'n', the third is n(n-1)/2!, the fourth is n(n-1)(n-2)/3! etc.

This may be on your formula book.
7. (Original post by notnek;[url="tel:62561237")
62561237[/url]]It will make your life easier if you remember the pattern of the coefficients so you don't have to derive each one.

E.g. for the simple expansion (1+x)^n, the second term coefficient is 'n', the third is n(n-1)/2!, the fourth is n(n-1)(n-2)/3! etc.

This may be on your formula book.
Thank you again for your help

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