Question
Expand (x-1/2y)^5
Answers
Answered by
GPT 3.5
To expand (x - \frac{1}{2}y)^5, we can use the binomial theorem, which states:
(x + y)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^2 + \binom{n}{3}x^{n-3}y^3 + ... + \binom{n}{n-1}xy^{n-1} + \binom{n}{n}y^n
In this case, n = 5, x = x, and y = -\frac{1}{2}y. Plugging these values into the binomial theorem formula, we have:
(x - \frac{1}{2}y)^5 = \binom{5}{0}x^5 + \binom{5}{1}x^4(-\frac{1}{2}y) + \binom{5}{2}x^3(-\frac{1}{2}y)^2 + \binom{5}{3}x^2(-\frac{1}{2}y)^3 + \binom{5}{4}x(-\frac{1}{2}y)^4 + \binom{5}{5}(-\frac{1}{2}y)^5
Now, let's calculate each term:
\binom{5}{0} = 1
\binom{5}{1} = 5
\binom{5}{2} = 10
\binom{5}{3} = 10
\binom{5}{4} = 5
\binom{5}{5} = 1
Plugging these values back into the expansion, we have:
(x - \frac{1}{2}y)^5 = 1 \cdot x^5 + 5 \cdot x^4 \cdot (-\frac{1}{2}y) + 10 \cdot x^3 \cdot (-\frac{1}{2}y)^2 + 10 \cdot x^2 \cdot (-\frac{1}{2}y)^3 + 5 \cdot x \cdot (-\frac{1}{2}y)^4 + 1 \cdot (-\frac{1}{2}y)^5
Simplifying each term:
x^5 - \frac{5}{2}x^4y + \frac{5}{2}x^3y^2 - \frac{5}{4}x^2y^3 + \frac{5}{8}xy^4 - \frac{1}{32}y^5
Therefore, the expansion of (x - \frac{1}{2}y)^5 is x^5 - \frac{5}{2}x^4y + \frac{5}{2}x^3y^2 - \frac{5}{4}x^2y^3 + \frac{5}{8}xy^4 - \frac{1}{32}y^5.
(x + y)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^2 + \binom{n}{3}x^{n-3}y^3 + ... + \binom{n}{n-1}xy^{n-1} + \binom{n}{n}y^n
In this case, n = 5, x = x, and y = -\frac{1}{2}y. Plugging these values into the binomial theorem formula, we have:
(x - \frac{1}{2}y)^5 = \binom{5}{0}x^5 + \binom{5}{1}x^4(-\frac{1}{2}y) + \binom{5}{2}x^3(-\frac{1}{2}y)^2 + \binom{5}{3}x^2(-\frac{1}{2}y)^3 + \binom{5}{4}x(-\frac{1}{2}y)^4 + \binom{5}{5}(-\frac{1}{2}y)^5
Now, let's calculate each term:
\binom{5}{0} = 1
\binom{5}{1} = 5
\binom{5}{2} = 10
\binom{5}{3} = 10
\binom{5}{4} = 5
\binom{5}{5} = 1
Plugging these values back into the expansion, we have:
(x - \frac{1}{2}y)^5 = 1 \cdot x^5 + 5 \cdot x^4 \cdot (-\frac{1}{2}y) + 10 \cdot x^3 \cdot (-\frac{1}{2}y)^2 + 10 \cdot x^2 \cdot (-\frac{1}{2}y)^3 + 5 \cdot x \cdot (-\frac{1}{2}y)^4 + 1 \cdot (-\frac{1}{2}y)^5
Simplifying each term:
x^5 - \frac{5}{2}x^4y + \frac{5}{2}x^3y^2 - \frac{5}{4}x^2y^3 + \frac{5}{8}xy^4 - \frac{1}{32}y^5
Therefore, the expansion of (x - \frac{1}{2}y)^5 is x^5 - \frac{5}{2}x^4y + \frac{5}{2}x^3y^2 - \frac{5}{4}x^2y^3 + \frac{5}{8}xy^4 - \frac{1}{32}y^5.