the first three terms in the expansion of (1+ay)to the power of n are 1, 12y, and 68y to the power of 2.
Evaluate a and n. us the factor that (1+ay)©ú= 1+nay+n(n-1) (ay)2+¡¦
2
Your second sentence/equation has many errors. You probably meant to write
" USE the FACT that
(1+ay)^n = 1 + nay +n(n-1)(ay)^2)/2 + ... "
This is called a Taylor series.
In this case, by comparing terms in an expansion in a power series of ay,
you may be able to solve for both a and n
To evaluate the values of a and n in the expansion of (1+ay)^n, we can use the given information about the first three terms.
The general form of the binomial expansion is given by the formula:
(1 + ay)^(n) = C(n,0)(ay)^0 + C(n,1)(ay)^1 + C(n,2)(ay)^2 + ...
Here, C(n, r) represents the binomial coefficient, which is given by the formula:
C(n, r) = n! / (r!(n-r)!)
Using the given information, we can substitute the values of the first three terms in the expansion:
First term: C(n, 0)(ay)^0 = 1 --> C(n, 0) = 1
Second term: C(n, 1)(ay)^1 = 12y --> C(n, 1) = 12
Third term: C(n, 2)(ay)^2 = 68y^2 --> C(n, 2) = 68
Now, let's evaluate the value of a:
C(n, 1) = 12 --> n! / (1!(n-1)!) = 12 --> n! / ((n-1)!) = 12
Since n! includes the factors of (n-1)!, the equation becomes:
n = 12
Now, let's evaluate the value of n:
C(n, 2) = 68 --> n! / (2!(n-2)!) = 68 --> n! / (2!(n-2)!) = 68
Simplifying this equation:
n(n-1) / 2 = 68 --> n(n-1) = 136
Now we can solve for n using trial and error:
n = 12 --> 12(12-1) = 132, which is less than 136
n = 13 --> 13(13-1) = 156, which is greater than 136
Thus, n = 13.
Now that we know n = 13, we can solve for a:
n = 12 --> 12! / (1!11!) = 12 --> 12! = 12(11!)
12! = 12! --> 11! = 1
Since 11! = 1, we can conclude that 12 = 12*a, which means a = 1.
Therefore, the values of a and n in the expansion (1+ay)^n are a = 1 and n = 13.