How do you use the binomial theorem to approximate #(2.99)^12#?

1 Answer
George C.
Sep 4, 2017

#2.99^12 ~~ 510568.79#

Explanation:

Note that:

#(2.99)^12 = (3-0.01)^12#

is of the form #(a+b)^n#, with #a=3#, #b=0.01# and #n=12#.

The binomial theorem tells us that:

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

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

Instead of calculating these binomial coefficients directly, we can pick them out from the row of Pascal's triangle that starts #1, 12,#...

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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#

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