math

expand (1/a+√a)^5

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asked by louise
  1. Use the binomial theorem,
    (X+Y)^n
    =(n,0)X^n+(n,1)X^(n-1)Y+....+(n,n-1)XY^(n-1)+(n,n)Y^n
    where
    (n,i) is the binomial coefficient
    =n!/[(n-i)!i!]

    For
    n=5,
    X=1/a
    Y=sqrt(a)
    we get
    (1/a+sqrt(a))^5
    =1/a^5 + (5/a^4)sqrt(a) + (10/a^3)a + (10/a^2)a^(3/2) + (5/a)a^2 + a^(5/2)
    which simplifies to
    1/a^5 + (5/a^(7/2) + 10/a^2 + 10/sqrt(a) + 5a + a^(5/2)

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