How do you use the binomial theorem to approximate #(2.99)^12#?
1 Answer
Sep 4, 2017
Explanation:
Note that:
#(2.99)^12 = (3-0.01)^12#
is of the form
The binomial theorem tells us that:
#(a+b)^n = sum_(k=0)^n ((n),(k)) a^(n-k) b^k#
where
Instead of calculating these binomial coefficients directly, we can pick them out from the row of Pascal's triangle that starts
That is:
#1,12,66,220,495,792,924,792,495,220,66,12,1#
Hence:
#2.99^12 = (3-0.01)^12#
#color(white)(2.99^12) ~~ 3^12-12(3^11)(0.01)+66(3^10)(0.01)^2-220(3^9)(0.01)^3+495(3^8)(0.01)^4#
#color(white)(2.99^12) = 531441-0.12(177147)+0.0066(59049)-0.000220(19683)+0.00000495(6561)#
#color(white)(2.99^12) = 531441-21257.64+389.7234-4.330260+0.03247695#
#color(white)(2.99^12) ~~ 510568.79#