Asked by Rethabile Milisi
find the first four terms using the binomial theorem
1/(sqrt(x^2 - (3/y))
1/(sqrt(x^2 - (3/y))
Answers
Answered by
Reiny
1/(sqrt(x^2 - (3/y))
= 1/(x^2 - 3/y)^(1/2)
= (x^2 - 3/y)^(-1/2)
= (x^2)^(-1/2) + (-1/2)(x^2)^(-3/2)(-3/y) + (-1/2)(-3/2)/2! (x^2)^(-5/2)(-3/y)2 + (-1/2)(-3/2)(-5/2)(-7/2)(1/3!)(x^2)^(-7/2) (-3/y)^3 + ...
= 1/x + 3/(2x^3 y) + 27/(8x^5 y^2) + 135/(16x^7 y^3) + ...
check my algebra
= 1/(x^2 - 3/y)^(1/2)
= (x^2 - 3/y)^(-1/2)
= (x^2)^(-1/2) + (-1/2)(x^2)^(-3/2)(-3/y) + (-1/2)(-3/2)/2! (x^2)^(-5/2)(-3/y)2 + (-1/2)(-3/2)(-5/2)(-7/2)(1/3!)(x^2)^(-7/2) (-3/y)^3 + ...
= 1/x + 3/(2x^3 y) + 27/(8x^5 y^2) + 135/(16x^7 y^3) + ...
check my algebra
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