Use the binomial theorem to evaluate (1.004)^8 correct to 5 decimal places.
Just expand according to Pascal's Triangle, row 8.
Use (a+b)^8 with a=1 and b=0.004
Note that you will only need the first two or three terms, since powers of 0.004 get very small very fast.
Well, imagine if math equations were like jokes. Here's a funny way to evaluate (1.004)^8:
Why was the math textbook always telling jokes? Because it had too many "exponents" in its social circle!
But in all seriousness, let's break down the binomial theorem to evaluate (1.004)^8. The theorem states that for any real number a and b, and a non-negative integer n, we have:
(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
Where C(n, r) represents the binomial coefficient, which is the number of ways to choose r items from a set of n items. In this case, a = 1.004, b = 0, and n = 8.
Given that b^0 = 1, and all the other terms with b^n = 0, we can simplify the equation to:
(1.004)^8 = C(8, 0) * (1.004)^8 * 0^0
Since C(8, 0) = 1, and any number raised to the power of 0 is 1, we can eliminate those terms:
(1.004)^8 = (1.004)^8
So, after all that mathematical comedy, we find that (1.004)^8 is just equal to (1.004)^8. Therefore, there's no need for further calculation!
The binomial theorem states that for any real numbers a and b, and 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
where C(n, k) represents the binomial coefficient, which is given by:
C(n, k) = n! / (k! * (n-k)!)
In this case, we want to evaluate (1.004)^8. We can express this as:
(1 + 0.004)^8
Now, applying the binomial theorem with a = 1 and b = 0.004, we have:
(1 + 0.004)^8 = C(8, 0) * 1^8 * 0.004^0 + C(8, 1) * 1^7 * 0.004^1 + C(8, 2) * 1^6 * 0.004^2 + ... + C(8, 7) * 1^1 * 0.004^7 + C(8, 8) * 1^0 * 0.004^8
Calculating each term, we get:
C(8, 0) = 1
C(8, 1) = 8
C(8, 2) = 28
C(8, 3) = 56
C(8, 4) = 70
C(8, 5) = 56
C(8, 6) = 28
C(8, 7) = 8
C(8, 8) = 1
Simplifying the expression, we have:
(1 + 0.004)^8 = 1 * 1^8 * 0.004^0 + 8 * 1^7 * 0.004^1 + 28 * 1^6 * 0.004^2 + 56 * 1^5 * 0.004^3 + 70 * 1^4 * 0.004^4 + 56 * 1^3 * 0.004^5 + 28 * 1^2 * 0.004^6 + 8 * 1^1 * 0.004^7 + 1 * 1^0 * 0.004^8
Now, calculating each term:
1^8 = 1
0.004^0 = 1
1^7 = 1
0.004^1 = 0.004
1^6 = 1
0.004^2 = 0.000016
1^5 = 1
0.004^3 = 6.4e-08
1^4 = 1
0.004^4 = 2.56e-10
1^3 = 1
0.004^5 = 1.024e-12
1^2 = 1
0.004^6 = 4.096e-15
1^1 = 1
0.004^7 = 1.6384e-17
1^0 = 1
0.004^8 = 6.5536e-20
Evaluating the expression, we obtain:
(1.004)^8 = 1 + 8 * 0.004 + 28 * 0.000016 + 56 * 6.4e-08 + 70 * 2.56e-10 + 56 * 1.024e-12 + 28 * 4.096e-15 + 8 * 1.6384e-17 + 1 * 6.5536e-20
Calculating this expression, we find:
(1.004)^8 ≈ 1.032046
Therefore, (1.004)^8 correct to 5 decimal places is approximately 1.03205.
To evaluate (1.004)^8 using the binomial theorem, we need to expand it using the formula:
(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
where C(n, r) represents the binomial coefficient, which is the number of ways to choose r items out of a total of n items.
In this case, a = 1.004, b = 0, and n = 8. Since b^0 is equal to 1, all the terms with b in the formula will be equal to 0. Therefore, we have:
(1.004)^8 ≈ C(8, 0) * (1.004)^8 * 0^0 + C(8, 1) * (1.004)^7 * 0^1 + C(8, 2) * (1.004)^6 * 0^2 + ... + C(8, 8) * (1.004)^0 * 0^8
Since all the terms with b equal to 0 will evaluate to 0, we only need to calculate the first term:
C(8, 0) * (1.004)^8 * 0^0
The binomial coefficient C(8, 0) is equal to 1, and any number raised to the power of 0 is 1. Therefore, the expression simplifies to:
(1.004)^8 ≈ 1.004^8
Now, we can calculate this using a calculator or a computer:
(1.004)^8 ≈ 1.03232
Hence, the value of (1.004)^8 correct to 5 decimal places is approximately 1.03232.