Question

write down in full the binomial expansion of (2-1/2x) rise to power 5 and simplify the coefficieng.(1) find the sum of the coefficient of x rise to power 3, x rise to power 4 and x rise to power 5 in the expansion.(2) use your expansion in (1) to calculate the value of (1.999) rise to power 5 correct your answer to 2singn figure.​

Answer 1

• 32 -40x +20x^2-5x^3 +5/8x^4 -x^5/32
• -4 13/32
• 31.92 (accurate to 3 decimal places); 32 to 2 significant figures

Explanation:

a) The full expansion of a binomial to the 5th power is ...

`(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5`

For the given binomial, a=2, b=-1/2x, so the expansion is ...

`(2-(x)/(2))^5=2^5-5(2^4)(x)/(2)+10(2^3)(x^2)/(2^2)-10(2^2)(x^3)/(2^3)+5(2)(x^4)/(2^4)-(x^5)/(2^5)\\\\=\boxed{32-40x+20x^2-5x^3+(5)/(8)x^4-(x^5)/(32)}`

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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)