Find the first four terms of the expansion of sqrt(1-3x). Hence evaluate √0.7 correct to 4 d.p
The first four terms of the expansion of $\sqrt{1-3x}$ can be found using the binomial theorem:
$$\sqrt{1-3x}=\sum_{n=0}^{\infty}\binom{1/2}{n}(-3x)^n$$
The first four terms of this expansion are:
$$\begin{align}\sqrt{1-3x}&=1-\frac{3x}{2}+\frac{9x^2}{8}-\frac{27x^3}{16}+\cdots\\
&=1-\frac{3x}{2}+\frac{9x^2}{8}-\frac{27x^3}{16}+O(x^4)\end{align}$$
To evaluate $\sqrt{0.7}$ correct to 4 decimal places, we can substitute $x=0.1$ into the above expansion since $0.7=1-0.3$. Therefore,
$$\begin{align}\sqrt{0.7}&=\sqrt{1-0.3}\\
&\approx 1-\frac{3(0.1)}{2}+\frac{9(0.1)^2}{8}-\frac{27(0.1)^3}{16}\\
&= 0.8367\end{align}$$
Therefore, $\sqrt{0.7}$ correct to 4 decimal places is 0.8367.
nice answer, bot -- but TOTALLY unfreadable
√(1-3x) = (1-3x)^(1/2)
= 1 - 3/2 x - 9/8 x^2 - 27/16 x^3 - ...
so now just plug and chug with x = 0.1
To find the first four terms of the expansion of √(1-3x), we can use the binomial expansion formula:
(1 + x)^n = C(n,0) * x^0 + C(n,1) * x^1 + C(n,2) * x^2 + C(n,3) * x^3 + ...
where C(n,k) represents the binomial coefficient, given by:
C(n,k) = n! / (k! * (n-k)!)
In this case, we have √(1-3x) which can be rewritten as (1-3x)^(1/2). Applying the binomial expansion formula, we get:
(1-3x)^(1/2) = C(1/2,0) * (3x)^0 + C(1/2,1) * (3x)^1 + C(1/2,2) * (3x)^2 + C(1/2,3) * (3x)^3 + ...
Now let's find the first four terms:
Term 1: C(1/2,0) * (3x)^0
= 1 * 1
= 1
Term 2: C(1/2,1) * (3x)^1
= (1/2) * 3x
= (3/2)x
Term 3: C(1/2,2) * (3x)^2
= (1/2 * -1/2) * (3)^2 * x^2
= (-1/4) * 9x^2
= (-9/4)x^2
Term 4: C(1/2,3) * (3x)^3
= (1/2 * -1/2 * -3/2) * (3)^3 * x^3
= (-1/8) * 27x^3
= (-27/8)x^3
Therefore, the first four terms of the expansion of √(1-3x) are:
1, (3/2)x, (-9/4)x^2, (-27/8)x^3
Now, to evaluate √0.7 correct to 4 decimal places, we substitute x = 0.7 into the expansion:
√0.7 ≈ 1 + (3/2)(0.7) + (-9/4)(0.7)^2 + (-27/8)(0.7)^3
Calculating this expression gives:
√0.7 ≈ 1 + 1.05 + (-0.735) + (-0.191)
= 1 + 1.05 - 0.735 - 0.191
= 1.114
Therefore, √0.7 ≈ 1.114 (correct to 4 decimal places).
To find the expansion of sqrt(1-3x), we can use the binomial theorem. The binomial theorem states that for a binomial expression (a + b)^n, the terms in the expanded form can be found using the formula:
C(n, k) * a^(n-k) * b^k
Where C(n, k) is the binomial coefficient, given by:
C(n, k) = n! / (k! * (n-k)!)
For the expression sqrt(1-3x), we can rewrite it as (1 + (-3x))^0.5
Using the binomial theorem, the first four terms can be found by expanding (1 + (-3x))^0.5 up to the fourth power (n=0 to n=3):
Term 1 (k=0): C(0, 0) * 1^0 * (-3x)^0 = 1 * 1 * 1 = 1
Term 2 (k=1): C(1, 1) * 1^0 * (-3x)^1 = 1 * 1 * (-3x) = -3x
Term 3 (k=2): C(2, 2) * 1^0 * (-3x)^2 = 1 * 1 * 9x^2 = 9x^2
Term 4 (k=3): C(3, 3) * 1^0 * (-3x)^3 = 1 * 1 * (-27x^3) = -27x^3
Hence, the first four terms of the expansion of sqrt(1-3x) are: 1, -3x, 9x^2, -27x^3
To evaluate sqrt(0.7) correct to 4 decimal places, we can use the expansion we found and substitute x with 0.7:
sqrt(0.7) = 1 - 3(0.7) + 9(0.7^2) - 27(0.7^3)
Calculating this expression, we get:
sqrt(0.7) = 1 - 2.1 + 4.83 - 11.025
Therefore, evaluating sqrt(0.7) correct to 4 decimal places, we get approximately 1.7050.