Review question

# How are the sums of $n$, $2n$ and $3n$ terms connected here? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8493

## Solution

1. If $S_1$, $S_2$ and $S_3$ are the sums of $n$, $2n$ and $3n$ terms of an arithmetic progression, show that $S_3=3(S_2-S_1)$.

Let our arithmetic progression be $u_1, u_2, u_3, ...$, or alternatively $a, a+d, a + 2d, ...$.

If we now take the sum over the first $n$ values, we find $\sum_{m=1}^{n} u_m = \dfrac{n}{2}(2a+(n-1)d).$

This is the formula we need. So for our sequence,

\begin{align*} S_1&=\dfrac {n}{2}(2a + (n-1)d),\\ S_2&=n(2a + (2n-1)d),\\ S_3&=\dfrac{3n}{2}(2a + (3n-1)d). \end{align*}

Thus $3(S_2-S_1) = 3n\left[2a+2nd-d-\left(a+\dfrac{nd}{2}-\dfrac{d}{2}\right)\right] = 3n\left[a+\dfrac{3nd}{2}-\dfrac{d}{2}\right] = \dfrac{3n}{2}(2a+(3n-1)d)= S_3$.

1. Find the sum of $n$ terms of the series $1+\frac{3x}{(1+2x^2)}+\frac{9x^2}{(1+2x^2)^2}+\frac{27x^3}{(1+2x^2)^3}+....$

This is a geometric series, with first term $a = 1$, and with common ratio $r = \dfrac{3x}{1+2x^2}$.

The formula for the sum of the first $n$ terms of a geometric progression is

$S_n = \dfrac{a(r^n-1)}{r-1} = \dfrac{a(1-r^n)}{1-r}.$

So here we have $S_n=\frac{\left(\dfrac{3x}{1+2x^2}\right)^n-1}{\dfrac{3x}{1+2x^2}-1}=\frac{\dfrac{(3x)^n-(1+2x^2)^n}{(1+2x^2)^n}}{\dfrac{3x-(1+2x^2)}{1+2x^2}}= \frac{(3x)^n -(1+2x^2)^n}{(1+2x^2)^{n-1}(1-2x)(x-1)}.$

For what values of $x$ does this series have a sum to infinity?

A geometric series diverges if and only if $\big \vert r \big \vert \geq 1.$ So when is $\big|\dfrac{3x}{1+2x^2}\big| \ge 1$ true?

Note that $1 + 2x^2 > 0$ for all real values of $x$, so we need to solve $\big|3x\big| \geq 1+2x^2$.

Let’s look at the case $x>0$; this gives us $2x^2-3x+1\geq 0$ to solve, which is true if and only if $(2x-1)(x-1) \geq 0$, or $\frac{1}{2}\le x \le 1$.

Now looking at the case $x<0$; we need to solve $2x^2+3x+1\geq 0$, which is true if and only if $(2x+1)(x+1) \geq 0$, or $-1 \leq x \leq -\frac{1}{2}$.

Hence we find the series to diverge for $\frac{1}{2}\le \big \vert x \big \vert \le 1$, and it will converge for all other values of $x$.