find term independent of y in expansion of,[x^4÷y^3+y^2÷2x]^10
If you consider that LCM(2,3)=6
a = x^4/y^3
b = y^2/2x
You want terms of the form
a^2b^3 or a^4b^6, ...
Since we have a 10th power expansion, the exponents must add to 10, so we are looking for a^4b^6. That is, the 5th term:
C(10,4) a^4 b^6
= 210 x^16/y^12 y^12/64x^6
= (105/32) x^10
Well, let's take a closer look at the expression and find the term that does not contain the variable "y."
The expression is:
[(x^4/y^3) + (y^2/2x)]^10
To find the term that does not involve "y," we want the term where all the y's cancel each other out when we expand the expression. This will happen when the powers of y in the numerator and denominator are equal. In this case, the y^3 in the denominator will match with the y^3 in the numerator, leaving us with:
x^4^10/y^3^10
Simplifying further, we get:
x^40/y^30
So, the term independent of y in the expansion is x^40.
To find the term independent of y in the expansion of [(x^4/y^3) + (y^2/2x)]^10, we can use the Binomial Theorem.
The Binomial Theorem states that the expansion of (a + b)^n can be calculated using the formula:
C(n, k) * a^(n-k) * b^k
where C(n, k) represents the binomial coefficient and can be calculated using the formula:
C(n, k) = n! / (k! * (n-k)!)
In our case, a = (x^4/y^3) and b = (y^2/2x), and we want to find the term where the exponent of y is 0.
The exponent of y in each term will be 10 - k (since we need n - k y terms to get an exponent of 0).
To find the term where the exponent of y is 0, we need (10 - k) to be equal to 0. Solving this equation gives us k = 10.
So, the term independent of y will be:
C(10, 10) * ((x^4/y^3)^(10-10)) * ((y^2/2x)^10)
Since ((x^4/y^3)^(10-10)) and ((y^2/2x)^10) are both equal to 1, we can simplify further:
C(10, 10) * 1 * 1
C(10, 10) = 10! / (10! * 0!) = 1
Therefore, the term independent of y in the expansion is simply 1.
In other words, the constant term in the expansion is 1.
To find the term independent of y in the expansion of [x^4/y^3 + y^2/2x]^10, we can use the binomial theorem.
The binomial theorem states that for any real numbers a and b, and any positive integer n, the expansion of the binomial (a + b)^n can be calculated using the formula:
(a + b)^n = C(n, 0)*a^n*b^0 + C(n, 1)*a^(n-1)*b^1 + C(n, 2)*a^(n-2)*b^2 + ... + C(n, n-1)*a^1*b^(n-1) + C(n, n)*a^0*b^n
Where C(n, k) represents the binomial coefficient. It is defined as:
C(n, k) = n! / (k!(n-k)!)
In our case, the given expression is [x^4/y^3 + y^2/2x]^10. Let's expand this using the binomial theorem.
The first term in the expansion will have a negative exponent of y because it contains the factor (x^4/y^3). To get rid of the negative exponent of y, we need to rewrite it as y^(-3).
Similarly, the second term y^2/2x can be rewritten as (1/2)yx^(-1). This is done to ensure that the exponents of y will remain positive.
Now, our expression becomes:
[(x^4/y^3) + (1/2)yx^(-1)]^10
Using the binomial theorem formula, the term independent of y will be of the form (a^m)(b^n), where m and n are the powers of a and b respectively.
In our case, a = x^4/y^3 and b = (1/2)yx^(-1).
To have a term independent of y, the exponent of y in (a^m)(b^n) must be 0. So, we need to find the values of m and n that satisfy m(-3) + n = 0.
Simplifying the equation, we get -3m + n = 0.
Now, let's find the values of m and n that satisfy this equation:
n = 3m
Since both m and n must be non-negative integers, we can start with m = 0 and find the corresponding value of n.
When m = 0, n = 3(0) = 0.
So, the term independent of y in the expansion of [x^4/y^3 + y^2/2x]^10 will be (x^4)^0 [(1/2)y(x^(-1))^0] = (1/2)x^0y^0 = 1.
Therefore, the term independent of y is 1.