Anonymous
Dec 11, 2023
By making use of binomial expansion determine the value of(2.002)^4 to four decimal places?
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Dec 11, 2023
To find the value of (2.002)^4 using binomial expansion, we can use the formula:
(1 + x)^n = 1 + nx + (n(n-1)x^2)/2! + (n(n-1)(n-2)x^3)/3! + ...
In this case, x = 0.002 and n = 4. Plugging these values into the formula, we get:
(1 + 0.002)^4 = 1 + 4(0.002) + (4(4-1)(0.002)^2)/2! + (4(4-1)(4-2)(0.002)^3)/3! + ...
Simplifying each term, we have:
(1 + 0.008 + (4(3)(0.000004))/2 + (4(3)(2)(0.000000008))/6 + ...
Calculating each term, we get:
(1 + 0.008 + 0.000024 + 0.000000032 + ...)
Adding up the terms, we find:
1 + 0.008 + 0.000024 + 0.000000032 ≈ 1.008024
Therefore, (2.002)^4 is approximately 1.008024 to four decimal places.
(1 + x)^n = 1 + nx + (n(n-1)x^2)/2! + (n(n-1)(n-2)x^3)/3! + ...
In this case, x = 0.002 and n = 4. Plugging these values into the formula, we get:
(1 + 0.002)^4 = 1 + 4(0.002) + (4(4-1)(0.002)^2)/2! + (4(4-1)(4-2)(0.002)^3)/3! + ...
Simplifying each term, we have:
(1 + 0.008 + (4(3)(0.000004))/2 + (4(3)(2)(0.000000008))/6 + ...
Calculating each term, we get:
(1 + 0.008 + 0.000024 + 0.000000032 + ...)
Adding up the terms, we find:
1 + 0.008 + 0.000024 + 0.000000032 ≈ 1.008024
Therefore, (2.002)^4 is approximately 1.008024 to four decimal places.
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