Binomial Questions? (1 Viewer)

[renny 123]

Q1. In the expansion of (3+4x)^n the coefficients of x^2 and x^3 are in the ratio of 3:4 find the value of n?​

Q2. If (1+px)^n= 1+15x+90x^2+..., find the values of p and n?



Thanks in advanced

 

[tommykins]

Q1.

coefficient of x^2
= nC2 . 3^(n-2) . (4)^2
= [n! . 3^(n-2) . (4)^2] / [2! . (n-2)!]

coefficient of x^3
= nC3 . 3^(n-3) . (4)^3
= [n! . 3^(n-3) . (4)^3] / [3! . (n-3)!]

{[n! . 3^(n-2) . (4)^2] / [2! . (n-2)!]} / {[n! . 3^(n-3) . (4)^3] / [3! . (n-3)!]} = 3/4

[n! . 3^(n-2) . (4)^2 . 3! . (n-3)!] / [n! . 3^(n-3) . (4)^3 . 2! . (n-2)!] = 3/4

18 / [8 . (n-2)] = 3/4

n = 5

Q2.
Coeff of x is 15

so nC1.p = 15
nC1 = n
so np = 15

repeat for x^2 and simulatenous equations.

 

[renny 123]

For question 1. i got five as well.
The answer said four though!
thats why i was very confused.

 

[bored of sc]

Q1. In the expansion of (3+4x)^n the coefficients of x^2 and x^3 are in the ratio of 3:4 find the value of n?​

Q2. If (1+px)^n= 1+15x+90x^2+..., find the values of p and n?



Thanks in advanced

1) T4/T3 = (n-3+1)/3 * 4/3 = 4/3
n-2 = 3
n = 5

2) T3/T2 = (n-2+1)/2 * p = 90/15
n-1 = 12/p
n = 12/p+1
= (12+p)/p #

T2 = nC1.1^n-1.(px)¹ = 15
= n.1.p for values of n and p
np = 15
(12+p)/p * p = 15
12+p = 15
p = 3
sub into #
n = 15/3
n = 5

Check:
(1+3x)⁵
= 1+5C1.1⁴.(3x)¹+5C2.1³.(3x)²
= 1+15x+90x²...

Yay!