Summation of series — practice questions

First, we split the summation into parts based on the standard formulae:

$S_n = \sum_{r=1}^{n} (r^2 - 3r + 2) = \sum_{r=1}^{n} r^2 - 3\sum_{r=1}^{n} r + \sum_{r=1}^{n} 2$

Now, substitute the standard formulae for each part:

$S_n = \frac{1}{6}n(n+1)(2n+1) - 3 \left( \frac{1}{2}n(n+1) \right) + 2n$

To combine these, we take out a common factor of $n$. It is also helpful to put everything over a common denominator of 6.

$S_n = \frac{n}{6} [ (n+1)(2n+1) - 9(n+1) + 12 ]$

Expand the brackets inside the square brackets:

$S_n = \frac{n}{6} [ (2n^2 + 3n + 1) - (9n + 9) + 12 ]$

Simplify the expression inside the square brackets:

$S_n = \frac{n}{6} [ 2n^2 + 3n + 1 - 9n - 9 + 12 ] = \frac{n}{6} [ 2n^2 - 6n + 4 ]$

We can factor out a 2 from the bracket:

$S_n = \frac{2n}{6} [ n^2 - 3n + 2 ] = \frac{n}{3} [ n^2 - 3n + 2 ]$

Finally, factorise the quadratic expression:

$S_n = \frac{1}{3}n(n-1)(n-2)$

i) Let $\frac{2}{(2r-1)(2r+1)} \equiv \frac{A}{2r-1} + \frac{B}{2r+1}$.

Multiplying by the denominator gives $2 \equiv A(2r+1) + B(2r-1)$.

Let $r = \frac{1}{2}$: $2 = A(1+1) + B(0) \implies 2 = 2A \implies A=1$.

Let $r = -\frac{1}{2}$: $2 = A(0) + B(-1-1) \implies 2 = -2B \implies B=-1$.

So, $\frac{2}{(2r-1)(2r+1)} = \frac{1}{2r-1} - \frac{1}{2r+1}$.

ii) We have found that the general term $u_r$ is in the form $f(r) - f(r+1)$ where $f(r) = \frac{1}{2r-1}$.

Let's write out the terms of the sum $S_n = \sum_{r=1}^{n} \left( \frac{1}{2r-1} - \frac{1}{2r+1} \right)$:

$r=1: \quad (\frac{1}{1} - \frac{1}{3})$

$r=2: \quad + (\frac{1}{3} - \frac{1}{5})$

$r=3: \quad + (\frac{1}{5} - \frac{1}{7})$

$\dots$

$r=n: \quad + (\frac{1}{2n-1} - \frac{1}{2n+1})$

By inspection, the $-\frac{1}{3}$ from the first term cancels with the $+\frac{1}{3}$ from the second, and so on. This is a telescoping sum. The only terms that do not cancel are the first term, $\frac{1}{1}$, and the last term, $-\frac{1}{2n+1}$.

Therefore, $S_n = 1 - \frac{1}{2n+1}$.

iii) The sum to infinity, $S_{\infty}$, is the limit of $S_n$ as $n \to \infty$.

$S_{\infty} = \lim_{n \to \infty} \left( 1 - \frac{1}{2n+1} \right)$

As $n \to \infty$, the denominator $2n+1 \to \infty$, so the fraction $\frac{1}{2n+1} \to 0$.

$S_{\infty} = 1 - 0 = 1$.