Find the first four terms of the expansion of sqrt(1-3x). Hence evaluate √0.7 correct to 4 d.p

Answer 1

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.

Answer 2

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