Find the value of x for which thr middle term in the expansion of ( 1 + x ) ^ 20 in ascending powers of x is the geometric mean of the 9th and the 12th terms.
the middle term of the expansion is term(10)
term(10) = C(20,9)(1)^11 x^9
= 167960x^9
the 9th term is
C(20,8) x^8 = 125970x^8
the 12 term is C(20,11) x^11 = 167960x^11
geometric mean = √(125970(167960)x^19)
167960x^9 = √(125970(167960)x^19)
167960x^9 = √(125970(167960))*x^9 √x
167960 = √(125970(167960) √x
square both sides
167960^2 = 125970(167960)x
I got this to reduce to x = 4/3
check:
middle term = 2236940.819 , using my calculator
term9 = 1258279.211
term12 = 3976783.676
G-mean = 2236940.819
yeahhhh!!!
Hmmm. Since there are 21 terms in the expansion, the 11th term is the middle one.
Adjust the numbers accordingly.
To find the middle term in the expansion of (1 + x)^20, we can use the binomial theorem. According to the binomial theorem, the middle term will be of the form:
C(n, k) * a^(n-k) * b^k,
where n is the power of the binomial, a is the first term (1), b is the second term (x), and k is the middle term.
In this case, n = 20, a = 1, and b = x.
Let's find the middle term using the formula:
k = (n+1)/2.
Plugging in the values, we have:
k = (20+1)/2 = 21/2 = 10.5.
However, since binomial coefficients are integers, the middle term must have a whole number exponent. Hence, we need to find the two middle terms: the 10th term (with exponent 10) and the 11th term (with exponent 11).
Now, let's find the 9th and 12th terms as well:
For the 9th term (with exponent 9), we have:
C(20, 9) * (1)^(20-9) * x^9 = C(20, 9) * x^9.
For the 12th term (with exponent 12), we have:
C(20, 12) * (1)^(20-12) * x^12 = C(20, 12) * x^12.
The geometric mean of the 9th and 12th terms is given by:
√(term9 * term12).
Let's substitute the values:
√((C(20, 9) * x^9) * (C(20, 12) * x^12)).
Now, we need to find the value of x for which this expression is equal to the middle term.
Let's substitute the values of k and calculate the middle term:
Middle term = C(20, 10) * (1)^(20-10) * x^10.
So, the equation to solve is:
√((C(20, 9) * x^9) * (C(20, 12) * x^12)) = C(20, 10) * (1)^(20-10) * x^10.
Simplifying further:
√((C(20, 9) * C(20, 12)) * x^(9+12)) = C(20, 10) * x^10.
Canceling out the common x terms:
√((C(20, 9) * C(20, 12)) * x^21) = C(20, 10) * x^10.
Taking the square root of both sides:
√(C(20, 9) * C(20, 12)) * x^(21/2) = C(20, 10) * x^10.
Now, we can cancel out the x terms by equating the powers:
21/2 = 10.
This simplifies to:
21 = 20.
Since this equation is not consistent, there is no value of x for which the middle term is the geometric mean of the 9th and 12th terms in the expansion of (1 + x)^20.