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
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 + ...
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?
go with Steve's solution
To find the values of p, q, and n, we need to analyze the given expression for the binomial expansion.
The expression (p + x)^n represents a binomial expansion of the form:
(p + x)^n = C(n, 0)p^n + C(n, 1)p^(n-1)x + C(n, 2)p^(n-2)x^2 + ... + C(n, n-1)px^(n-1) + C(n, n)x^n
where C(n, k) represents the binomial coefficient, given by the formula:
C(n, k) = n! / (k!(n-k)!)
From the given expression, we can equate the coefficients of x in both sides to find the values of p, q, and n.
Comparing the terms with x in the given expression:
1. The term with x has a coefficient of 6q, which corresponds to the term C(n, 1)p^(n-1)x in the binomial expansion. Therefore, 6q = C(n, 1)p^(n-1).
2. The term with x^2 has a coefficient of 5qx^2, which corresponds to the term C(n, 2)p^(n-2)x^2 in the binomial expansion. Therefore, 5q = C(n, 2)p^(n-2).
Since p ≠ 0, we can divide these two equations to eliminate p:
(6q) / (5q) = [C(n, 1)p^(n-1)] / [C(n, 2)p^(n-2)]
Simplifying:
6 / 5 = [n! / (1!(n-1)!)] / [n! / (2!(n-2)!)]
6 / 5 = (n! / (n-1)!) * ((2!(n-2)!) / n!)
6 / 5 = (n(n-1) / 1) * (1 / (2(n-2)))
6 / 5 = n(n-1) / (2n - 4)
6(2n - 4) = 5n(n-1)
12n - 24 = 5n^2 - 5n
5n^2 - 17n + 24 = 0
Now we can solve this quadratic equation for n. By factoring or using the quadratic formula, we find that n = 3 or n = 8/5. However, since n represents the exponent in a binomial expansion, it should be a positive integer. Therefore, n = 3.
Substituting n = 3 back into the original equations, we can find the values of p and q.
From equation 1, 6q = C(3, 1)p^2
6q = 3p^2
2q = p^2
From equation 2, 5q = C(3, 2)p
5q = 3p
q = (3/5)p
Substituting q = (3/5)p into the equation 2q = p^2, we get:
2(3/5)p = p^2
6p/5 = p^2
6p = 5p^2
Dividing both sides by p (since p ≠ 0), we get:
6/5 = 5p
6 = 25p
p = 6/25
Substituting p = 6/25 back into q = (3/5)p, we get:
q = (3/5)(6/25)
q = 18/125
Therefore, the values of p, q, and n are p = 6/25, q = 18/125, and n = 3.