Some binomials (1 Viewer)

[currysauce]

1.

When (3+2x)^n is expanded in increasing powers of x, it is found that the coeffs. of x^5 and x^6 have the same values. Find the value for n and show that the two coefficients mentioned are greater than all other coeffs. in the expansion.

2.

Write the binomial expansion of (1+x)^n in ascending powers of x. DONE. THen, find the value of n such that the coeff. of x^4 is twice the coeff. of x^3.

3.

Expand (1 + (1/x))^3 in ascending powers of x. DONE. THen, if x - (1/x) = 1, find the value of x^3 - (1/x^3).

thanks!

 

[香港!]

"1.

When (3+2x)^n is expanded in increasing powers of x, it is found that the coeffs. of x^5 and x^6 have the same values. Find the value for n and show that the two coefficients mentioned are greater than all other coeffs. in the expansion."

general term of the expansion (3+2x)^n:
nCk (3)^(n-k)(2x)^k=nCk (3)^(n-k) (2)^k x^k

"the coeffs. of x^5 and x^6 have the same values"
so nC5 (3)^(n-5) (2)^5=nC6 (3)^(n-6) (2)^6
nC5 (3)=nC6 (2)
3n!\(5!(n-5)! ) =2n!\(6!(n-6)!)

3\(5!(n-5)! )=2\(6!(n-6)!)
5!(n-5)! \ 3=6!(n-6)!) \ 2
(n-5)! \ 3=3 (n-6)!
(n-5)!=9 (n-6)!
(n-5)=9
n=14

then do the usual thing for getting the greatest coefficient

 

[香港!]

"2.

Write the binomial expansion of (1+x)^n in ascending powers of x. DONE. THen, find the value of n such that the coeff. of x^4 is twice the coeff. of x^3."

general term for (1+x)^n:
nCk x^k

"coeff. of x^4 is twice the coeff. of x^3"
so nC4=2* nC3
n!\(4!(n-4)!)=2n!\(3!(n-3)!)
4!(n-4)!=3!(n-3)!\2
8=(n-3)
n=11

 

[香港!]

"3.

Expand (1 + (1/x))^3 in ascending powers of x. DONE. THen, if x - (1/x) = 1, find the value of x^3 - (1/x^3).

thanks!"

(1 + (1/x))³=1+3(1\x)+3(1\x)²+(1\x)³

"x - (1/x) = 1"
so x=1+(1\x)
x³=(1+(1\x))³
x³-(1\x³)=1+3(1\x)+3(1\x)²

 

[currysauce]

the answer to 3 is 4

can u try and get it?

 

[香港!]

the answer to 3 is 4

can u try and get it?

x - (1/x) = 1
x²-1=x
x²-x-1=0
x=(1+-sqrt(1+4))\2
x=(1+sqrt5)\2


x³-(1\x³)=1+3(1\x)+3(1\x)²
=1+6\(1+sqrt5)+12\(1+sqrt5)²
=4 by calculator

 

[currysauce]

ok thanks, is that the only way to do that quesiton?

 

[shafqat]

x - 1/x = 1
Cubing both sides, x^3 - 3x + 3/x - 1/x^3 = 1
so x^3 - 1/x^3 = 4

 

[currysauce]

this what u mean

x^3 - 1/x^3 -3(x - 1/x) = 1

x^3 - 1/x^3 -3(1) = 1

x^3 - 1/x^3 = 4 ???

 

[shafqat]

x - 1/x = 1
Cubing both sides, x^3 - 3x + 3/x - 1/x^3 = 1
x^3 - 1/x^3 - 3(x - 1/x) = 1
So x^3 - 1/x^3 - 3 = 1
so x^3 - 1/x^3 = 4

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