Solve Binomial Theorem: Find Term Independent of x

[thornluke]

Homework Statement
The method of Binomial expansion is useful because you can avoid expanding large expressions:
Q: Find the term indepedent of x in the expansion of (2+x)[2x+(1/x)]⁵

The attempt at a solution:
"For this to produce a term independent of x, the expansion of [2x+(1/x)]⁵ must have a constant term or a term in x^-1...?
So the power of x is given by 5-2r. This cannot be zero for positive integer values of r. Hence the required coefficient is given by:
5 - 2r = -1
r = 3"

But why?! I do not understand that underlined statement.

 

[Saitama]

Homework Statement
The method of Binomial expansion is useful because you can avoid expanding large expressions:
Q: Find the term indepedent of x in the expansion of (2+x)[2x+(1/x)]⁵

The attempt at a solution:
"For this to produce a term independent of x, the expansion of [2x+(1/x)]⁵ must have a constant term or a term in x^-1...?
So the power of x is given by 5-2r. This cannot be zero for positive integer values of r. Hence the required coefficient is given by:
5 - 2r = -1
r = 3"

But why?! I do not understand that underlined statement.

What would be happen if it becomes zero? See there's a term (2+x) also present.

 

[thornluke]

What would be happen if it becomes zero? See there's a term (2+x) also present.

So...the expansion of [2x+(1/x)]⁵ must have a constant term or a term in x^-1...

Constant term means the term with no x right?

But why can the term independent of x be in x^-1 too?

 

[Infinitum]

But why can the term independent of x be in x^-1 too?

Because the term already has a x^-1! It cannot be independent of x if it already has a x

To explain the question, you need a term independent of x. You get them independent of x, if a constant term gets multiplied with another constant term, OR a term with x^-1 is multiplied with a term with x. But there is no constant term(not involving x) in the expansion of [2x+(1/x)]⁵ as explained. So the only possibility remains, that x^-1 gets multiplied with x.

 

[thornluke]

Because the term already has a x^-1! It cannot be independent of x if it already has a x

I am so lost here.. why am I so terrible at maths

 

[Saitama]

Constant term means the term with no x right?

Yes!

We need to find the constant term in expansion of (2+x)(2x+(1/x))⁵, not in (2x+(1/x))⁵. So when x in (2+x) multiplies by a term having x^-1 with it, we get a constant term as $x*\frac{1}{x}=1.$.

 

[thornluke]

Yes!

We need to find the constant term in expansion of (2+x)(2x+(1/x))⁵, not in (2x+(1/x))⁵. So when x in (2+x) multiplies by a term having x^-1 with it, we get a constant term as $x*\frac{1}{x}=1.$.

Ahh.. I think I'm starting get the grasp of this.. one more problem!
Expand (2+x)⁵ and hence find 1.9⁵
= 32 + 80x+ 80x² + 40x³ + 10x⁴ + x⁵

Why can 1.9⁵ be considered to be x= -0.1 in the above expansion?

 

[Saitama]

Ahh.. I think I'm starting get the grasp of this.. one more problem!
Expand (2+x)⁵ and hence find 1.9⁵
= 32 + 80x+ 80x² + 40x³ + 10x⁴ + x⁵

Why can 1.9⁵ be considered to be x= -0.1 in the above expansion?

What is (2+(-0.1))⁵=?

 

[thornluke]

What is (2+(-0.1))⁵=?

Thanks very much. Can't believe I can't even think simple.

 

[cool_jessica]

its binomial theorem.
use the formula t(r+1)=nCr*x^n-r*a6r
x=x^2 and a=-1/2x (substitution of the question variables)
n=power=3
let the term independent of x be t(r+1)=x^0=1
therefore after substitution the formula would be
Y*x^0=nCr*(x^2)^3-r*(-1*x^ -1)^r
therefore power of x=6-2r-r
but in the term power of x is supposed to be 0
0=6-3r
6=3r
r=2.
therefore the term is r+1
2+1
=3..
therefore the term independent of x is the third term.