Find the first four terms in the binomial expansion of (1+x)^(1/5) and state the range of values of x for which this expansion is valid.Hence approximate (31)^(1/5) correct to four decimal places.
If you are studying this, you must know
(1+x)^n = 1 + nx + n(n-1)(x^2)/2! + ...
valid only for |x| < 1
now let's apply it to your problem of 31^(1/5)
= (32 - 1)^(1/5)
= 32^(1/5)(1 - 1/32)^(1/5)
= 2 (1 - 1/32)^(1/5))
= 2( 1 + (1/5)(-1/32) + (1/5)(-4/5)/2 (-1/32)^2 + (1/5)(-4/5)(-9/5)/6 (-1/32)^3 + .... )
= 2(1 -.00625 + .000078125 - .000001465 + ...)
= 2(.9938266...)
= 1.98765..
using my calculator to find 32^(1/5)
I got 1.98734
Good work, but you had to use a different function. Poorly set problem...
To find the first four terms in the binomial expansion of (1+x)^(1/5), we can use the binomial theorem:
(1+x)^(1/5) = C(1/5,0) * 1^(1/5) * x^0 + C(1/5,1) * 1^(1/5-1) * x^1 + C(1/5,2) * 1^(1/5-2) * x^2 + C(1/5,3) * 1^(1/5-3) * x^3 + ...
where C(n, r) represents the binomial coefficient "n choose r".
The general formula for the binomial coefficient C(n, r) is given by:
C(n, r) = n! / (r!(n-r)!)
Using this formula, we can find the coefficients for the first four terms:
C(1/5, 0) = 1
C(1/5, 1) = (1/5)
C(1/5, 2) = (1/5)(1/5 - 1) / (2!)
C(1/5, 3) = (1/5)(1/5 - 1)(1/5 - 2) / (3!)
Simplifying the expressions, we get:
C(1/5, 0) = 1
C(1/5, 1) = (1/5)
C(1/5, 2) = -(1/25) / 2 = -1/50
C(1/5, 3) = (1/125) / 6 = 1/750
So the first four terms in the binomial expansion of (1+x)^(1/5) are:
1 + (1/5)x - (1/50)x^2 + (1/750)x^3
The range of values of x for which this expansion is valid is when |x| < 1, as this ensures convergence within the radius of convergence of the binomial series.
Now, we can approximate (31)^(1/5) to four decimal places:
Plugging in x = 31 into the binomial expansion, we get:
(31)^(1/5) = 1 + (1/5)(31) - (1/50)(31)^2 + (1/750)(31)^3
Calculating this expression:
(31)^(1/5) ≈ 1 + 6.2 - 1.53 + 0.72
(31)^(1/5) ≈ 7.39
Therefore, (31)^(1/5) is approximately equal to 7.39, correct to four decimal places.
To find the first four terms in the binomial expansion of (1+x)^(1/5), we can use the binomial theorem. The binomial theorem states that for any real number n and any real number x, the binomial expansion of (1+x)^n can be calculated using the formula:
(1 + x)^n = C(n, 0) * 1^n * x^0 + C(n, 1) * 1^(n-1) * x^1 + C(n, 2) * 1^(n-2) * x^2 + ... + C(n, n-1) * 1^1 * x^(n-1) + C(n, n) * 1^0 * x^n
where C(n, r) represents the binomial coefficient, given by the formula:
C(n, r) = n! / (r! * (n - r)!)
In this case, n = 1/5 and we need the first four terms, so we can substitute these values into the formula:
Term 1: C(1/5, 0) * 1^(1/5) * x^0
Term 2: C(1/5, 1) * 1^(1/5 - 1) * x^1
Term 3: C(1/5, 2) * 1^(1/5 - 2) * x^2
Term 4: C(1/5, 3) * 1^(1/5 - 3) * x^3
Let's calculate these terms step by step:
Term 1: C(1/5, 0) * 1^(1/5) * x^0 = 1 * 1 * 1 = 1
Term 2: C(1/5, 1) * 1^(1/5 - 1) * x^1 = (1/5) * 1 * x = 1/5 * x
Term 3: C(1/5, 2) * 1^(1/5 - 2) * x^2 = (1/5) * (-4/5) * x^2 = -4/25 * x^2
Term 4: C(1/5, 3) * 1^(1/5 - 3) * x^3 = (1/5) * (-4/5) * (-2/5) * x^3 = 8/125 * x^3
So, the first four terms in the binomial expansion of (1+x)^(1/5) are:
Term 1: 1
Term 2: 1/5 * x
Term 3: -4/25 * x^2
Term 4: 8/125 * x^3
Now, let's determine the range of values of x for which this expansion is valid. The binomial expansion is valid when |x| is less than 1. In this case, |x| < 1 means -1 < x < 1.
Finally, to approximate (31)^(1/5) correct to four decimal places, we can substitute x = 31 into the binomial expansion expression and calculate it:
(31)^(1/5) ≈ Term 1 + Term 2 + Term 3 + Term 4
(31)^(1/5) ≈ 1 + 1/5 * 31 + (-4/25) * 31^2 + 8/125 * 31^3
Using a calculator, we can simplify this expression to get:
(31)^(1/5) ≈ 1 + 6.2 + (-96.58) + 198.688
(31)^(1/5) ≈ 109.282
Therefore, (31)^(1/5) is approximately equal to 109.282, correct to four decimal places.