How do we find the value of(1.01)^5 approximately up to three decimals,only with the use of the Binomial theorem?
(1 + .01)^5 = 1^5 + 5(1)^4*.01 + (20/2)(1)^3*.01^2 etc
Thank you!
You are welcome.
To find the approximate value of (1.01)^5 using the Binomial theorem, we can use the following formula:
(1 + x)^n ≈ 1 + nx + (n(n-1)x^2 / 2!) + (n(n-1)(n-2)x^3 / 3!) + ...
In this formula, we substitute x with the value we want to calculate (in this case, x = 0.01) and n with the power we want to raise it to (in this case, n = 5).
So let's substitute:
(1.01)^5 ≈ 1 + 5(0.01) + (5(5-1)(0.01)^2 / 2!) + (5(5-1)(5-2)(0.01)^3 / 3!) + ...
Now let's calculate these terms step by step:
1 + 5(0.01) = 1.05
5(5-1)(0.01)^2 / 2! = 10(0.01)^2 / 2 = 10(0.0001) / 2 = 0.0005
5(5-1)(5-2)(0.01)^3 / 3! = 10(0.01)^3 / 6 = 10(0.000001) / 6 = 0.000001666...
Now, let's substitute these values back into the original formula to get the approximate value of (1.01)^5:
(1.01)^5 ≈ 1 + 1.05 + 0.0005 + 0.000001666...
By adding up these values, we can approximate (1.01)^5 up to three decimal places.
1 + 1.05 + 0.0005 = 2.0505
Therefore, (1.01)^5 is approximately equal to 2.0505 up to three decimal places, using the Binomial theorem.