Question

Use the binomial theorem to expand the expression :

(3x + y)^5 and simplify.
(b) find the middle term in the expansion of
(1/x+√x)^4 and simplify your unswer.
(c) determine the coefficient of x^11 in the expansion of (x^2 +1/x)^10, simplify your answer.

Answer 1

a)
`(3x+y)^5=243x^5+405x^4y+270x^3y^2+90x^2y^3+15xy^4+y^5`.

b) The middle term in the expansion is
`(6)/(x)`.

c) The coefficient of
`x^(11)` is 120.

Explanation:

Remember that the binomial theorem say that
`(x+y)^n=\sum_(k=0)^(n) \binom{n}{k}x^(n-k)y^(k)`

a)
`(3x+y)^5=\sum_(k=0)^5\binom{5}{k}3^(n-k)x^(n-k)y^k`

Expanding we have that

`\binom{5}{0}3^5x^5+\binom{5}{1}3^4x^4y+\binom{5}{2}3^3x^3y^2+\binom{5}{3}3^2x^2y^3+\binom{5}{4}3xy^4+\binom{5}{5}y^5`

symplifying,

`(3x+y)^5=243x^5+405x^4y+270x^3y^2+90x^2y^3+15xy^4+y^5`.

b) The middle term in the expansion of
`((1)/(x) +√(x))^4=\sum_(k=0)^(4)\binom{4}{k}(1)/(x^(4-k))x^{(k)/(2)}` correspond to k=2. Then
`\binom{4}{2}(1)/(x^2)x^{(2)/(2)}=(6)/(x)`.

c)
`(x^2+(1)/(x))^(10)=\sum_(k=0)^(10)\binom{10}{k}x^(2(10-k))(1)/(x^k)=\sum_(k=0)^(10)\binom{10}{k}x^(20-2k)(1)/(x^k)=\sum_(k=0)^(10)\binom{10}{k}x^(20-3k)`

Since we need that 11=20-3k, then k=3.

Then the coefficient of
`x^(11)` is
`\binom{10}{3}=120`