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