binomial help please (from fitzpatrick) (1 Viewer)

kr73114 · Mar 3, 2011

1) question 30 of 29a in fitzpatrick 3u, soz i cant write it in in this post.

2) In the expansion of (2+3x)^n the coefficients of x^3 and x^4 are in the ratio 8:15. Find n

3) Find n if the coefficents of the 2nd, 3rd and 4th terms in the expansion of (1+x)^n are successive terms of an arithmetic sequence.

ohexploitable · Mar 3, 2011

I don't have the book so I can't do (1).

For (2),

$\,\frac{\binom{n}{3}2^{n-3}3^3}{\binom{n}{4}2^{n-4}3^4}=\frac{8}{15}$

$\frac{\frac{n!}{3!(n-3)!}\times2\times2^{n-4}}{\frac{n!}{4!(n-4)!}\times2^{n-4}}=\frac85$

$2\times\frac{3!\times4\times(n-4)!}{3!(n-4)!(n-3)}=\frac85$

$\frac{8}{n-3}=\frac85$

$n=8$

4025808 · Mar 3, 2011

For (1)

since for somewhat reason I can't use latex, it'll take so long and it'll be hard to read so here's the general picture

terms cancel out when you expand with the binomial theorem, there are ones that don't cancel out.

you should end up with 10(x-1)^2 + 20(x-1) +2
simplifying gives 10x^2 -8

and for others, here's the question

-> simplify (sqrt[x-1] +1)^5 - (sqrt[x -1] -1)^5

ohexploitable · Mar 3, 2011

For (3),

$\binom{n}{4}-\binom{n}{3}=\binom{n}{3}-\binom{n}{2}$

Then simplify using $\binom{n}{r}=\frac{n!}{r!(n-r)!}$

Get equal denominators and clean up.

hscishard · Mar 3, 2011

kr73114 said:
1) question 30 of 29a in fitzpatrick 3u, soz i cant write it in in this post.

It might help to expand it all using the binomial theorem.
You should get [2 x 5C1 (x-1)^2] + [2 x 5C3(x-1)] + 2 x 5C5
Then it would simplify to 10x^2 - 8. Just becareful when you play with the minus signs

4025808 · Mar 3, 2011

ohexploitable said:
$\binom{n}{4}-\binom{n}{3}=\binom{n}{3}-\binom{n}{2}$

btw that's wrong
it should be
$\binom{n}{3}-\binom{n}{2}=\binom{n}{2}-\binom{n}{1}$
because they are asking for the 2nd, 3rd and 4th coefficients, not the degrees of x

to OP
simplify the above and then you should get n =7

ohexploitable · Mar 3, 2011

4025808 said:
btw that's wrong
it should be
$\binom{n}{3}-\binom{n}{2}=\binom{n}{2}-\binom{n}{1}$
because they are asking for the 2nd, 3rd and 4th coefficients, not the degrees of x

to OP
simplify the above and then you should get n =7

lol yeah, my bad :S

hscishard · Mar 3, 2011

4025808 said:
btw that's wrong
it should be
$\binom{n}{3}-\binom{n}{2}=\binom{n}{2}-\binom{n}{1}$
because they are asking for the 2nd, 3rd and 4th coefficients, not the degrees of x

to OP
simplify the above and then you should get n =7

To make simplification easier, divide both sides by n! and multiply both sides by (n-3)! Tnen you'll only have a simple quadratic

kr73114 · Mar 3, 2011

thanks everybody

shaon0 · Mar 3, 2011

A=(sqrt(x-1)+1))^5-(sqrt(x-1)-1)^5
Let u=sqrt(x-1):
A=(u+1)^5-(u-1)^5
=(u^5+5u^4+10u^3+10u^2+5u^2+1)-(u^5-5u^4+10u^3-10u^2+5u-1)
=10u^4+20u^2+2 [All odd degrees cancel]
=2(5u^4+10u^2+1)
=2(5(x-1)^2+10(x-1)+1)
=2(5x^2-4)

binomial help please (from fitzpatrick) (1 Viewer)

kr73114

Member

ohexploitable

hey

4025808

Well-Known Member

ohexploitable

hey

hscishard

Active Member

4025808

Well-Known Member

ohexploitable

hey

hscishard

Active Member

kr73114

Member

shaon0

...

Users Who Are Viewing This Thread (Users: 0, Guests: 1)