Asked by louise
expand (1/a+√a)^5
Answers
Answered by
MathMate
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)
(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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