Use your expansion to evaluate 0.997^6, correct to four decimal place
We can use the binomial expansion formula to expand (1-0.003)^6 as follows:
(1-0.003)^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 substitute 0.997 for (1-0.003) and simplify:
0.997^6 = 1 - 6(0.003)(0.997) + 15(0.003)^2(0.997)^2 - 20(0.003)^3(0.997)^3 + 15(0.003)^4(0.997)^4 - 6(0.003)^5(0.997)^5 + (0.003)^6(0.997)^6
Using a calculator, we get:
0.997^6 ≈ 0.9821
Therefore, 0.997^6, correct to four decimal places, is approximately 0.9821.
To evaluate 0.997^6 using the expansion method, we can use the binomial expansion. The expansion for (1 + x)^n, where n is a positive integer, is given by:
(1 + x)^n = 1 + nx + (n(n-1)/2!)(x^2) + (n(n-1)(n-2)/3!)(x^3) + ...
In this case, we have x = -0.003 (since 0.997 is less than 1) and n = 6. Therefore, we can substitute these values into the expansion to get:
(1 + (-0.003))^6 = 1 + 6(-0.003) + (6(6-1)/2!)((-0.003)^2) + (6(6-1)(6-2)/3!)((-0.003)^3) + ...
Simplifying this expression step-by-step, we get:
(1 + (-0.003))^6 = 1 - 0.018 + 0.054 + 0.008 + ...
Now, we can round this expression to four decimal places:
1 - 0.018 + 0.054 + 0.008 = 1.044
Therefore, 0.997^6, correct to four decimal places, is approximately 1.044.
To evaluate 0.997^6, we can use the expansion method. Let's expand (1 - 0.003)^6 using the binomial expansion formula:
(1 - 0.003)^6 = 1^6 - 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 simplify and calculate each term:
1^6 = 1
6*(0.003) = 0.018
15*(0.003)^2 = 0.000135
20*(0.003)^3 = 0.00000243
15*(0.003)^4 = 0.00000002025
6*(0.003)^5 = 0.000000000054
(0.003)^6 = 0.000000000000729
Adding all these terms together, we get:
1 - 0.018 + 0.000135 - 0.00000243 + 0.00000002025 - 0.000000000054 + 0.000000000000729
Simplifying further, we get:
0.982161775375
To round this to four decimal places, we look at the fifth decimal place:
0.9821
Therefore, 0.997^6, correct to four decimal places, is approximately equal to 0.9821.