1a)write down and simplify all the terms of the
binomial expansion (1-x)6. (b) use your
expansion to evaluate 0.997^6 correct to 4d.p
(c) when 0.997^6 was computed by a student,
his result was 0.9817. Find the percentage
error correct to 3d.p.....
Plz help me show work thankx
expand the binomial. then, put .003 in for x. You wont have to go past the x^3 term...
binomial theorem:
(a + b)^n = sum[k=0,n][(n over k)a^(n-k)b^k], so
(1-x)^6=
1(6!/6!)1^6+ 6!/5! *1*x+6!/4!2! x^5 + ..... I think that will be far enough...
1-6*.003+15*.003^2+...= 0.982135 and lets do the next term to check
-6!/3!3! x^4=10*.003^4= 5.1E-10 so we have the answer...
0.982135
Now compare this to the "student's result".
as another check enter this into your google calculator (search window): .997^6=
a) To find the terms of the binomial expansion (1-x)^6, we can use the Binomial Theorem:
(1-x)^6 = C(6,0)(1)^6(-x)^0 + C(6,1)(1)^5(-x)^1 + C(6,2)(1)^4(-x)^2 + C(6,3)(1)^3(-x)^3 + C(6,4)(1)^2(-x)^4 + C(6,5)(1)^1(-x)^5 + C(6,6)(1)^0(-x)^6
Simplifying each term:
= 1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6
b) To evaluate 0.997^6, we substitute x = 0.003 into the expansion we found in part (a):
0.997^6 ≈ 1 - 6(0.003) + 15(0.003)^2 - 20(0.003)^3 + 15(0.003)^4 - 6(0.003)^5 + (0.003)^6
Now we can calculate this expression:
0.997^6 ≈ 0.9816
So, the answer to part (b) is 0.9816.
c) To find the percentage error, we can use the formula:
percentage error = (|exact value - approximate value| / exact value) * 100
Let's substitute the given values:
percentage error = (|0.997^6 - 0.9817| / 0.997^6) * 100
Calculating this:
percentage error = (|0.9816 - 0.9817| / 0.9816) * 100
= (0.0001 / 0.9816) * 100
≈ 0.0102%
Therefore, the percentage error correct to 3 decimal places is approximately 0.010%.
I don't understand
And there is no solution to the question
Sure! Let's break down the problem step by step:
(a) To find the terms of the binomial expansion (1-x)^6, we can use the binomial theorem. The binomial theorem states that for any positive integer n, the expansion of (a+b)^n can be written as:
(a+b)^n = C(n,0) * a^n * b^0 + C(n,1) * a^(n-1) * b^1 + C(n,2) * a^(n-2) * b^2 + ... + C(n,n-1) * a^1 * b^(n-1) + C(n,n) * a^0 * b^n
In this case, a = 1 and b = -x. We want to find the expansion of (1-x)^6, so n = 6. Let's calculate the terms:
C(6,0) * 1^6 * (-x)^0 = 1 * 1 * 1 = 1
C(6,1) * 1^5 * (-x)^1 = 6 * 1 * (-x) = -6x
C(6,2) * 1^4 * (-x)^2 = 15 * 1 * x^2 = 15x^2
C(6,3) * 1^3 * (-x)^3 = 20 * 1 * (-x)^3 = -20x^3
C(6,4) * 1^2 * (-x)^4 = 15 * 1 * x^4 = 15x^4
C(6,5) * 1^1 * (-x)^5 = 6 * 1 * (-x)^5 = -6x^5
C(6,6) * 1^0 * (-x)^6 = 1 * 1 * x^6 = x^6
So, the simplified terms of the binomial expansion (1-x)^6 are:
1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6.
(b) To evaluate 0.997^6 using the binomial expansion, we substitute x = 0.003 into the simplified terms we found earlier. Let's calculate:
0.997^6 ≈ 1 - 6(0.003) + 15(0.003)^2 - 20(0.003)^3 + 15(0.003)^4 - 6(0.003)^5 + (0.003)^6
Calculating each term:
1 - 0.018 + 0.00135 - 0.000027 - 1.0125e-07 + 9.67e-11 + 2.18e-13 ≈ 0.9817
(c) To find the percentage error, we compare the calculated result (0.9817) with the student's result (0.9817). The formula for percentage error is:
Percentage Error = [(Correct Value - Approximate Value) / Correct Value] * 100
Let's calculate the percentage error:
Percentage Error = [(0.9817 - 0.9817) / 0.9817] * 100 = 0%
Therefore, the percentage error is 0%, which means the student's result was accurate to the correct value of 0.9817.