Question

How do you use the binomial theorem to approximate `(1.98)^9`?

Answer 1

`1.98^9 ~~ 467.721`

Explanation:

By the binomial theorem:

`(a+b)^n = sum_(k=0)^n ((n),(k)) a^(n-k) b^k`

where `((n),(k)) = (n!)/((n-k)!k!)`

We could use this with `a=2` and `b=-0.02`, but it's probably cleaner to separate out the factor `2^9 = 512` and use `a=1` and `b=-0.01` instead.

We can find the binomial coefficients `((9),(k))` we need from the row of Pascal's triangle that starts `1, 9, 36,...`

So we find:

`1.98^9 = 2^9*(1-0.01)^9`

`color(white)(1.98^9) = 2^9*(1-9(0.01)+36(0.0001)-84(0.000001)+...)`

`color(white)(1.98^9) = 2^9*(1-0.09+0.0036-0.000084+...)`

`color(white)(1.98^9) ~~ 2^9*(1-0.09+0.0036-0.000083)`

`color(white)(1.98^9) ~~ 2^9*0.913517`

`color(white)(1.98^9) ~~ 2^8*1.827034`

`color(white)(1.98^9) ~~ 2^7*3.654068`

`color(white)(1.98^9) ~~ 2^6*7.308136`

`color(white)(1.98^9) ~~ 2^5*14.616272`

`color(white)(1.98^9) ~~ 2^4*29.232544`

`color(white)(1.98^9) ~~ 2^3*58.465088`

`color(white)(1.98^9) ~~ 2^2*116.930176`

`color(white)(1.98^9) ~~ 2*233.860352`

`color(white)(1.98^9) ~~ 467.721`