The coefficient of x^2 is 69 in the expansion of (1-ax)^24, where a>0. Find the coefficient of x^3.
(1-ax)^24
= 1 + C(24,1)(-ax) + C(24,2)(-ax)^2 + ...
so C(24,2)(a^2)x^2 = 69x^2
276a^2 = 69
a^2 = 69/276 = 1/4
a = ±1/2
if a = 1/2
term containing x^3 = C(24,3)(-x/2)^3
= -2024/8x^3 = -253x^3
if a = -1/2
term containing x^3 = C(24,3(x/2)^3 = 253x^3
thanks
To find the coefficient of \( x^3 \), we need to use the binomial theorem for the expansion \((1-ax)^{24}\).
The binomial theorem states that for any real number \( a \) and any positive integer \( n \), the expansion of \((1+ax)^n\) can be written as:
\[
(1+ax)^n = \binom{n}{0} a^0 x^0 + \binom{n}{1} a^1 x^1 + \binom{n}{2} a^2 x^2 + \ldots + \binom{n}{n} a^n x^n
\]
where \( \binom{n}{k} \) represents the binomial coefficient, and it can be calculated using the formula:
\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\]
Given that the coefficient of \( x^2 \) is 69, we can write the expansion as:
\[
(1-ax)^{24} = \binom{24}{0} a^0 x^0 + \binom{24}{1} a^1 x^1 + 69 x^2 + \binom{24}{3} a^3 x^3 + \ldots + \binom{24}{24} a^{24} x^{24}
\]
Since we know that the coefficient of \( x^2 \) is 69, we can equate it to the term \( \binom{24}{2} a^2 x^2 \):
\[
69 = \binom{24}{2} a^2
\]
Using the formula for the binomial coefficient, we can calculate \( \binom{24}{2} \):
\[
\binom{24}{2} = \frac{24!}{2!(24-2)!} = \frac{24 \times 23}{2} = 12 \times 23 = 276
\]
Now we can solve the equation for \( a^2 \):
\[
69 = 276a^2
\]
Dividing both sides by 276:
\[
\frac{69}{276} = a^2
\]
Simplifying:
\[
\frac{1}{4} = a^2
\]
Taking the square root of both sides:
\[
a = \frac{\sqrt{1}}{\sqrt{4}}
\]
Since \( a > 0 \), we can discard the negative square root:
\[
a = \frac{1}{2}
\]
Now that we know the value of \( a \), we can find the coefficient of \( x^3 \) by looking at the term \( \binom{24}{3} a^3 x^3 \):
\[
\binom{24}{3} = \frac{24!}{3!(24-3)!} = \frac{24 \times 23 \times 22}{3 \times 2 \times 1} = 24 \times 23 \times 4
\]
So the coefficient of \( x^3 \) is \( 24 \times 23 \times 4 \).
To find the coefficient of x^3 in the expansion of (1-ax)^24, we can use the binomial theorem. The general formula for the term in the expansion of (1-ax)^n is:
(n choose r) * (1^r) * ((-ax)^(n-r))
In this case, we are looking for the term with x^3, so we need to find the value of r that satisfies (24 choose r) * (1^r) * ((-ax)^(24-r)) = x^3.
The coefficient of x^2 is given as 69, which means that the term with x^2 has a coefficient of 69. Using the binomial theorem formula, we can set up the following equation:
(24 choose r) * (1^r) * ((-ax)^(24-r)) = x^2
Simplifying further, we have:
(24 choose r) * (-1)^(24-r) * (a^(24-r)) * x^(24-r) = x^2
Comparing the exponents of x on both sides of the equation, we get:
24 - r = 2
Solving for r, we find:
r = 22
Now that we know the value of r, we can substitute it into the original equation to find the coefficient of x^3:
(24 choose 22) * (-1)^2 * (a^2) * x^2 = x^3
Simplifying further, we have:
(24! / (22! * 2!)) * a^2 * x^2 = x^3
Dividing both sides of the equation by x^2, we get:
(24! / (22! * 2!)) * a^2 = x
Therefore, the coefficient of x^3 in the expansion of (1-ax)^24 is (24! / (22! * 2!)) * a^2.
Well, well, well, it seems we have some mathematical mischief going on! To find the coefficient of x^3 in the expansion of (1-ax)^24, we need to put our clown nose on and do some calculations.
Now, the coefficient of x^2 in the expansion of (1-ax)^24 is given to be 69. That means, when we expand (1-ax)^24, it looks something like this: ... + 69x^2 + ...
To find the coefficient of x^3, we need to think about how we can obtain an x^3 term when you multiply (1-ax) by itself 24 times. But let me tell you, multiplying (1-ax) twenty-four times is not exactly my idea of a fun time at the carnival!
But fear not! There's a shortcut called the Binomial Theorem that can save us from this circus act. It tells us that the coefficient of x^n in the expansion of (1+bx)^m is given by the binomial coefficient C(m, n) times b^n, where C(m, n) represents the number of ways to choose n items from a set of m items.
In this case, we want the coefficient of x^3, so we have n = 3. The exponent of x in (1-ax)^24 will be 3 when we choose the 'a' term 1 time from 24 terms. Mathematically, C(24, 1) represents this, and it equals 24.
So, the coefficient of x^3 will be 24 times the value of 'a' to the power of 1. Since we don't have the exact value of 'a' given, we can't determine the coefficient of x^3.
So, sorry to be the bearer of bad news, but this little clown can't find the coefficient you're looking for without knowing the value of 'a.'