# Expanding this binomial?

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Expand and simplify #(sqrt3 + 2)^5# . Give your answer in the from #a+bsqrt3# where a and b are whole, positive intergers.

Expand and simplify

##### 1 Answer

#### Explanation:

Let us look at what happens when

Using FOIL...

#(a+bsqrt(3))(sqrt(3)+2) = overbrace(a*sqrt(3))^"First"+overbrace(a*2)^"Outside"+overbrace(bsqrt(3)*sqrt(3))^"Inside"+overbrace(bsqrt(3)*2)^"Last"#

#color(white)((a+bsqrt(3))(sqrt(3)+2)) = (2a+3b)+(a+2b)sqrt(3)#

So starting with

#{ (a_(i+1) = 2a_i+3b_i), (b_(i+1)=a_i+2b_i) :}#

Writing

#(1, 0) -> (2, 1) -> (7, 4) -> (26, 15) -> (97, 56) -> (362, 209)#

So:

#(sqrt(3)+2)^5 = 362+209sqrt(3)#

**Observations**

What I notice about this is that the numbers:

#2/1, 7/4, 26/15, 97/56, 362/209#

are successively better rational approximations to

Why should this be?

Check the standard continued fraction expansion of

#sqrt(3) = [1;bar(1,2)] = 1+1/(1+1/(2+1/(1+1/(2+1/(1+1/(2+...))))))#

The partial sums are:

#1, color(blue)(2/1), 5/3, color(blue)(7/4), 19/11, color(blue)(26/15), 71/41, color(blue)(97/56), 265/153, color(blue)(362/209)#

The sequence of ratios we found are the (Pell equation satisfying) approximations:

#[1;1] = 2/1#

#[1;1,2,1] = 7/4#

#[1;1,2,1,2,1] = 26/15#

#[1;1,2,1,2,1,2,1] = 97/56#

#[1;1,2,1,2,1,2,1,2,1] = 362/209#

The rules we found for multiplying