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

1 Answer
Aug 23, 2017

Answer:

#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,...#

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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#