binomial help please (from fitzpatrick) (1 Viewer)

kr73114

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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.
 

4025808

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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
 

hscishard

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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
 

hscishard

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btw that's wrong
it should be

because they are asking for the 2nd, 3rd and 4th coefficients, not the degrees of x :p

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
 

shaon0

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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)
 

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