Question

A binomial expansion is such that

(p + x)^n = 3q + 6qx + 5qx² + ..

If p ≠ 0, find the values of

p, q and n
plz show step

Answer 1

well, just write the terms and equate powers:

(p+x)^n = p^n + np^(n-1) x + n(n-1)/2 p^(n-2) x^2 + ...

so,

p^n = 3q
np^n/p = 6q
n(n-1)/2 p^n/p^2 = 5q

n*3q/p = 6q
3n=6p
n=2p

n(n-1)/2*3q/p^2 = 5q
3n(n-1)=10p^2
3(2p)(2p-1)=10p^2
12p^2-6p=10p^2
2p(p-3)=0
p=0 or 1, but not 0, so p=3
n=2p=6
p^n=3q, so 3^6=3q, so q=3^5

(3 + x)^6 = 3^6 + 6*3^5*x + 6*3^4*x^2 + ...
= 3*3^5 + 6*3^5*x 6*5/2 * 3^4*x^2 + ...
= 3q + 6qx + 5qx^2 + ...

Answer 2

The first 3 terms in the expansion of
(p+x)^n
= p^n + n p^(n-1) x + n(n-1)/2 p^(n-2) x^2 + ..

so we have:
p^n = 3q **
n p^(n-1) x = 6qx ---> n p^(n-1) = 6q ***
n(n-1)/2 p^(n-2) x^2 = 5q x^2, or
n(n-1) p^(n-2) = 10q x^2 ****

** ÷ ***
p/n = 1/2 ---> n = 2p

*** ÷ ****
p/(n-2) = 3/5
5p = 3n - 6 , sub in n = 2p
5p = 6p - 6
p = 6 , and n = 12

in **
p^n = 3q
6^12 = 3q
q = 725594112 ???

better check that.

(6+x)^12
= 6^12 + 12(6^11) x + 12(11)/2 (6^10) x^2 + ...
= 3(725594112) + 12(36279056)x + 66(60433176)x^2 +
= 3(725594112) + 6(725594112)x + 5.5q

arrrghhh, where is my arithmetic error?

Answer 3

go with Steve's solution