Question

Expanding this binomial?

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

Answer 1

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

Explanation:

Let us look at what happens when `a+bsqrt(3)` is multiplied by `(sqrt(3)+2)`...

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_0=1` and `b_0=0`, we can apply these formulae `5` times:

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

Writing `(a_i, b_i)` as a pair, we find:

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

So:

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

`color(white)()`
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 `sqrt(3)`

Why should this be?

Check the standard continued fraction expansion of `sqrt(3)`:

`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 `a+bsqrt(3)` by `(sqrt(3)+2)` are essentially the same as evaluating two steps of this continued fraction.