If (1+px)^n = 1 + 15x +90x^2 +... , find the values of p and n.
Answer: p=3, n=5
Thanks for any help
expanding the binomial ... (1 + px)^n = nC0 1^n + nC1 1^(n-1) (px)^1 +
nC2 1^(n-2) (px)^2 + ...
2nd term ... n p = 15
3rd term ... [n (n - 1) / 2] p^2 = 90
dividing terms ... (n - 1) p = 12 ... n p - p = 12
subtracting equations ... p = 3
substitute back to find n
To find the values of p and n in the equation (1 + px)^n = 1 + 15x + 90x^2 + ..., we need to match the coefficients of the terms on both sides of the equation.
Let's start by expanding the left side of the equation using the binomial theorem:
(1 + px)^n = 1 + nx(px) + (n)(n-1)/2! (px)^2 + (n)(n-1)(n-2)/3! (px)^3 + ...
We can see that the coefficient of the term with x raised to the power of 0 on the left side is 1, which matches the right side.
The coefficient of the term with x raised to the power of 1 on the left side is nxp, which must be equal to 15x. Therefore, we can conclude that np = 15.
The coefficient of the term with x raised to the power of 2 on the left side is (n)(n-1)/2! (px)^2, which must be equal to 90x^2. Simplifying this, we get n(n-1)p^2/2 = 90.
Now we have two equations:
np = 15
n(n-1)p^2/2 = 90
We can solve these equations simultaneously for p and n.
From the first equation, np = 15, we can divide both sides by n to get p = 15/n.
Substituting this value of p into the second equation, we get n(n-1)(15/n)^2/2 = 90. Simplifying, we have n(n-1)(225/n^2)/2 = 90.
Next, we simplify further by canceling out common factors:
n(n-1)(225/2n^2) = 90
(n-1)(225/2n) = 90
Now, we can multiply both sides by 2n to eliminate the fraction:
(n-1)(225) = 180n
Expanding the left side:
225n - 225 = 180n
Rearranging the equation:
225n - 180n = 225
45n = 225
n = 5
Finally, substituting the value of n into np = 15:
p = 15/n = 15/5 = 3
Therefore, the values of p and n are p = 3 and n = 5.