Answer:
Step-by-step explanation:
a) The full expansion of a binomial to the 5th power is ...
[tex](a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5[/tex]
For the given binomial, a=2, b=-1/2x, so the expansion is ...
[tex](2-\dfrac{x}{2})^5=2^5-5(2^4)\dfrac{x}{2}+10(2^3)\dfrac{x^2}{2^2}-10(2^2)\dfrac{x^3}{2^3}+5(2)\dfrac{x^4}{2^4}-\dfrac{x^5}{2^5}\\\\=\boxed{32-40x+20x^2-5x^3+\dfrac{5}{8}x^4-\dfrac{x^5}{32}}[/tex]
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b) The sum of coefficients of x^3, x^4 and x^5 is ...
-5 +5/8 -1/32 = -4 13/32
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c) 1.999^5 = (2 -.001)^5 = (2 -0.002/2)^5
So, we can use the above expansion with x=.002. The result from part (b) tells us that the error from neglecting 3rd-power terms and higher will be on the order of 40×10^-9, far less than that necessary for the required accuracy.
1.999^5 ≈ 32 -.002(40 -.002(20)) = 32 -.002(39.96) = 32 -0.07992
1.999^5 ≈ 31.92
= 32 (to 2 significant figures)
