Series — practice questions

i) First, rewrite the expression in index form: $\sqrt{1-4x} = (1-4x)^{1/2}$. This is in the form $(1+y)^n$ with $y = -4x$ and $n = \frac{1}{2}$.

Using the binomial expansion formula $1 + ny + \frac{n(n-1)}{2!}y^2 + \dots$:

$= 1 + (\frac{1}{2})(-4x) + \frac{(\frac{1}{2})(\frac{1}{2}-1)}{2!}(-4x)^2 + \dots$

$= 1 - 2x + \frac{(\frac{1}{2})(-\frac{1}{2})}{2}(16x^2) + \dots$

$= 1 - 2x + \frac{-\frac{1}{4}}{2}(16x^2) + \dots$

$= 1 - 2x - \frac{1}{8}(16x^2) + \dots$

$= 1 - 2x - 2x^2 + \dots$

Let the first term be $a$ and the common ratio be $r$.

We are given: The second term is 12: $u_2 = ar = 12$ (Equation 1) The sum to infinity is 50: $S_\infty = \frac{a}{1-r} = 50$ (Equation 2)

From Equation 1, we can express $a$ in terms of $r$: $a = \frac{12}{r}$.

Substitute this expression for $a$ into Equation 2: $\frac{12/r}{1-r} = 50$

Now, solve for $r$: $12 = 50r(1-r)$ $12 = 50r - 50r^2$ $50r^2 - 50r + 12 = 0$ Divide by 2 to simplify: $25r^2 - 25r + 6 = 0$

This is a quadratic equation in $r$. We can factorise it: $(5r-2)(5r-3) = 0$ So, $r = \frac{2}{5}$ or $r = \frac{3}{5}$.

Both values of $r$ are positive and satisfy the condition $|r|<1$ for the sum to infinity to exist.

Case 1: $r = \frac{2}{5}$ $a = \frac{12}{r} = \frac{12}{2/5} = 12 \times \frac{5}{2} = 30$.

Case 2: $r = \frac{3}{5}$ $a = \frac{12}{r} = \frac{12}{3/5} = 12 \times \frac{5}{3} = 20$.

The question asks for the first term, and we have found two possible values.

The possible values for the first term are 20 and 30.