4.4 Sigma notation for sums — Practise with help

1. $1 + 2 + 3 + 4 + 5 = 15$.
2. $2(3x + 4) = 6x + 8$.
3. $\dfrac{30}{2} = 15$.

If question 1 was hard, add the numbers in pairs before you continue. The standard results in this section are built from sums like this one.

Substitute $r = 1, 2, 3, 4$ one at a time, and add the four terms you get.

The first term is $2(1) + 1 = 3$ and the second is $2(2) + 1 = 5$. Write down all four terms before you add them.

The sum has four terms, because the index runs from $1$ to $4$. Each term is larger than the one before by $2$.

Find the rule that produces the terms, then find the value of the index that gives the last term.

The $r$th multiple of $3$ is $3r$. Set $3r = 30$ to find the last value of $r$.

The terms are multiples of $3$, so the rule is $3r$. The last term is $30$, and $3r = 30$ gives $r = 10$. So

Checking the last term is the way to confirm the upper limit.

Substitute each value of $r$ and work out the square first, then subtract $1$.

The first two terms are $1^2 - 1 = 0$ and $2^2 - 1 = 3$. Continue to $r = 5$.

There are five terms. The first term is $0$, so the sum is not the sum of the squares $1 + 4 + 9 + 16 + 25 = 55$; it is five less.

Split the sum into $3\sum r$ plus the sum of the constant $2$. The constant $2$ is added once for each value of $r$.

$\sum_{r=1}^{10} r = \frac{10 \times 11}{2} = 55$, and the constant gives $10 \times 2 = 20$.

The sum of the constant is $10 \times 2$, not $2$, because there are ten terms.

Use $\sum_{r=1}^{n} r^2 = \dfrac{n(n+1)(2n+1)}{6}$ with $n = 6$.

Substitute $n = 6$: the numerator is $6 \times 7 \times 13$. Work that out before you divide by $6$.

Check with the terms: $1 + 4 + 9 + 16 + 25 + 36 = 91$. The formula and the list agree.

The sum does not start at $1$. Add the terms from $1$ to $8$, then take away the terms from $1$ to $2$.

$\sum_{r=1}^{8} r = \frac{8 \times 9}{2} = 36$ and $\sum_{r=1}^{2} r = 3$. The answer is the difference.

Check by adding: $3 + 4 + 5 + 6 + 7 + 8 = 33$. The adjustment works because the two sums share the same start, so their difference leaves the middle part.

The index runs over the month numbers, from $1$ to $5$. The summand is the income in month $n$.

Write $\sum_{n=1}^{5}600(0.8)^{n-1}$, then write out the five terms. Each one is $0.8$ times the one before.

Adding the brackets: $1 + 0.8 + 0.64 + 0.512 + 0.4096 = 3.3616$, so the total is $600 \times 3.3616 = 2016.96$. The appeal raises £2016.96 in the first five months. A geometric sum like this one has its own short formula, which Section 4.6 gives.