find five term of the expansion
(x+y/2)^1/3
show step plz
(x+y/2)^1/3
= x^(1/3) + (1/3)(x^(-2/3) (y/2) + (1/3)(-2/3)/2! (x^(-5/3) )(y/2)^2 + (1/3)(-2/3)(-5/3)/3! (x^(-5/3))(y/2)^3 + ....
= x^(1/3) + (1/6)(x^(-2/3)y - (1/36)(x^(-5/3)y^2 + (5/648)x^(-8/3)y^3 + ???
I left out the 5th term for you to finish.
I am sure you can see the pattern
1/0! x^(1/3) (y/2)^0
+ (1/3)/1! x^(-2/3) (y/2)^1
+ (1/3)(-2/3)/2! x^(-5/3)(y/2)^2
+ (1/3)(-2/3)(-5/3)/3! x^(-8/3)(y/2)^3
+ (1/3)(-2/3)(-5/3)(-8/3)/4! x^(-11/3)(y/2)^4
= x1/3
+ 1/6 x-2/3 y
- 1/36 x-5/3 y2
+ 5/648 x-8/3 y3
- 5/1944 x-11/3 y4
To find the five terms of the expansion of (x + y/2)^(1/3), we can use the binomial theorem. The formula for the binomial theorem is:
(x + y)^n = C(n, 0) * x^n * y^0 + C(n, 1) * x^(n-1) * y^1 + C(n, 2) * x^(n-2) * y^2 + ... + C(n, r) * x^(n-r) * y^r + ... + C(n, n) * x^0 * y^n
where C(n, r) represents the binomial coefficient, given by the formula:
C(n, r) = n! / (r! * (n-r)!)
Now, let's apply this formula to find the five terms of the expansion. In this case, n = 1/3.
First term:
C(1/3, 0) * (x)^1/3 * (y/2)^0 = 1 * (x)^1/3 * 1 = x^(1/3)
Second term:
C(1/3, 1) * (x)^(1/3 - 1) * (y/2)^1 = (1/3) * (x)^0 * (y/2)^1 = (1/3) * (y/2)
Third term:
C(1/3, 2) * (x)^(1/3 - 2) * (y/2)^2 = (1/3) * (1/3 - 1) * (x)^(-1/3) * (y/2)^2 = (1/3) * (-2/3) * (1/x) * (y/2)^2 = (-2/9) * (y^2 / 4x)
Fourth term:
C(1/3, 3) * (x)^(1/3 - 3) * (y/2)^3 = (1/3) * (1/3 - 1) * (1/3 - 2) * (x)^(-2/3) * (y/2)^3 = (1/3) * (-2/3) * (-5/3) * (1/x^2) * (y/2)^3 = (10/81) * (y^3 / 8x^2)
Fifth term:
C(1/3, 4) * (x)^(1/3 - 4) * (y/2)^4 = (1/3) * (1/3 - 1) * (1/3 - 2) * (1/3 - 3) * (x)^(-3/3) * (y/2)^4 = (1/3) * (-2/3) * (-5/3) * (-8/3) * (1/x^3) * (y/2)^4 = (-40/243) * (y^4 / 16x^3)
So, the five terms of the expansion of (x + y/2)^(1/3) are:
Term 1: x^(1/3)
Term 2: (1/3) * (y/2)
Term 3: (-2/9) * (y^2 / 4x)
Term 4: (10/81) * (y^3 / 8x^2)
Term 5: (-40/243) * (y^4 / 16x^3)