binomial therom (1 Viewer)

cos16mh · Feb 11, 2018

Can I have assistance for this question please

The expansion of (1+an)^n is given by 1-24x+252x^2-...Find values for a and n

Could you please explain how to do this thanks

integral95 · Feb 11, 2018

cos16mh said:
Can I have assistance for this question please

The expansion of (1+an)^n is given by 1-24x+252x^2-...Find values for a and n

Could you please explain how to do this thanks

I think you mistyped it and meant $(1+ax)^n$

integral95 · Feb 11, 2018

cos16mh said:
Can I have assistance for this question please

The expansion of (1+an)^n is given by 1-24x+252x^2-...Find values for a and n

Could you please explain how to do this thanks

Expanding the first 3 terms using Binomial thereom gives

$(1+ax)^n = 1 + \binom{n}{1} (ax) + \binom{n}{2}(ax)^2 + \dots = 1-24x+252x^2+ \dots$

Equate the co-efficient with last 2 terms to get a set of simultaneous equations to solve.

$\binom{n}{1} a = -24 \\ \\ \binom{n}{2}a^2 = 252$

I got n = 8 and a = -3

cos16mh · Feb 12, 2018

ThANKS
Could you explain how the simultaneous equations worked please

integral95 · Feb 12, 2018

cos16mh said:
ThANKS
Could you explain how the simultaneous equations worked please

Substitution method, make a the subject in the first equation.

$an = -24 \Rightarrow a = -\frac{24}{n}$

Sub that into the 2nd equation to get

$\frac{n!}{2!(n-2)!}(-\frac{24}{n})^2 = 252 \\ \\ \Rightarrow \frac{n(n-1)}{2} \frac{576}{n^2} = 252 \\ \\$

In case you're confused, this is because

$\frac{n!}{(n-2)!} = \frac{n(n-1)(n-2)!}{(n-2)!} = n(n-1)$

After simplifying and solving that equation you get n = 8 then sub it back to the first equation to get a = -3

cos16mh · Feb 15, 2018

Thank you so much

binomial therom (1 Viewer)

cos16mh

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