MATH1011 Chap.3 Sigma Notation and the Binomial Theorem
Sigma Notation and the Binomial Theorem
Reading a summation as an instruction
A capital sigma is an instruction, not an object. It says: take the counter written below it, run that counter from the value written below to the value written above, feed every whole number in between into the rule on the right, and add what comes out. Read it that way and the first two questions about any sum answer themselves. How many terms are there?
Upper limit minus lower limit, plus one. What does the counter mean once the adding is finished? Nothing whatever, which is why one sum written with i and another written with j can stand for the same number.
A constant sitting inside the sum is not idle.
It is delivered once for every value the counter takes, so adding 5 across twelve values contributes 60 rather than 5. A constant multiplying every term behaves in the opposite way, because it can be lifted out and applied once at the end.
Keeping those two apart is the single most common repair in this chapter.
Renaming, cutting and rescaling the index
Three edits leave a sum worth precisely what it was worth before. You may rename the counter, replacing it everywhere by a letter differing from the old one by a fixed whole number, provided both limits move by that same amount and the rule is rewritten too.
You may cut the range into two ranges that meet exactly, with no value counted twice and none skipped over. And you may pull a constant factor out in front.
Nearly every manipulation later in the course is one of those three, usually done so two sums acquire matching ranges and their terms can be paired off.
Four short results deserve to be known by sight rather than rebuilt: adding 1 across n values gives n; the first n whole numbers add to n(n + 1)/2; their squares add to n(n + 1)(2n + 1)/6; and the powers of x from 0 up to n add to (1 − x^(n+1))/(1 − x), which says nothing when x is 1. Verify the third on a tiny case: 1 + 4 + 9 + 16 is 30, and 4 × 5 × 9 divided by 6 is also 30.
Pi notation and the cancellation it invites
Swap the sigma for a capital pi and every rule above carries over with multiplication in place of addition, but two things change.
A constant factor repeated across n values becomes that constant raised to the power n, not that constant times n. And one factor equal to zero drags the whole product to zero however large the others may be. The reward is that many products collapse.
Multiply (k + 1)/k for k running from 1 to 5 and write the factors out: 2/1, then 3/2, then 4/3, then 5/4, then 6/5. Every top line is undone by the bottom line of the next factor, the middle material disappears, and the product is simply 6.
One coefficient without the whole expansion
A two-term bracket raised to the power n is n copies of that bracket in a row.
Multiplying them out means going along the row, taking one term from each copy, and adding every product you can build that way. A product holding k copies of the second term arises once for each choice of which k brackets handed it over, so the number standing in front of that product is a selection count from the previous chapter.
That is the entire content of the binomial formula, and it explains why the term in a squared, b cubed inside a fivefold bracket carries the coefficient 10: there are 10 ways of choosing which three of the five brackets supplied b.
Questions almost never want the expansion; they want one term of it, so describe a typical term first and compute last.
Take one term in terms of the counter, with any coefficient and minus sign locked inside their bracket before the power is applied. Collect every appearance of the variable into a single exponent. Equate that exponent with the power named in the question, using zero when the wanted term is the constant one, and solve for the counter. Only now is there any multiplying to do, and only one term of it.
Should the counter come out a fraction, or land outside its allowed range, you already have your answer: the expansion holds no such term.
What this chapter covers
- 01
Reading a sigma as a list of terms
- 02
Renaming the counter without changing the value
- 03
Splitting a range and lifting constants out
- 04
Standard closed forms worth knowing by sight
- 05
Pi notation and products that cancel
- 06
The binomial formula and one named coefficient
Worked example · free
A sum moved onto a new counter, and a single coefficient pulled out of a seventh power
- 2Rename the counter so the sum begins at 1, reading the new upper limit off the same substitution.
- 2Split the renamed sum into a scaled standard sum plus a constant delivered once per value of the counter.
- 2Evaluate both pieces, then audit the total against the first term, the last term and the number of terms.
- 2Write one general term of the expansion with the coefficient and the minus sign held inside their bracket.
- 2Collect the powers into a single exponent, solve for the counter, and evaluate that one term only.
Key terms
- Summation Index
- The summation index is the counter written beneath the sigma, taking each whole number from the lower limit to the upper limit in turn. It belongs to the sum alone, so replacing every appearance of it with another letter leaves the value untouched.
- Index Shift
- An index shift swaps the counter for a new one differing from it by a fixed whole number, then rewrites both limits and the rule to match. Its usual purpose is to force two sums onto one range so their terms can be paired off.
- Pi Notation
- Pi notation compresses a product the way sigma notation compresses a sum, using the same counter, the same pair of limits and one rule for each factor. A single factor of zero sends the whole product to zero, which has no parallel among sums.
- Telescoping Product
- A telescoping product is one in which each factor cancels part of the next, so that nearly everything disappears and only pieces of the first and last factors remain. Recognising one turns a long product into a short fraction with no formula needed.
- General Term
- The general term is the single expression, carrying one unknown counter, that stands for every term of an expansion at once. Writing it down with all numerical factors and signs enclosed is what lets one named coefficient be found without multiplying the bracket out.
- Pascal Recurrence
- The Pascal recurrence builds each binomial coefficient by adding the two coefficients standing immediately above it in the previous row. It follows from asking whether one particular object was among those chosen, and it is the quickest way to obtain a small coefficient by hand.
Sigma Notation and the Binomial Theorem FAQ
How do I move the starting index of a sum without changing its value?
Decide what the new starting value should be, note how far the counter has to travel, then move both limits and the rule by that same amount in that same direction. Confirm with three numbers afterwards: the first term, the last term and the count of terms must all come out as they did before the move.
Can I find one coefficient without expanding the whole bracket?
Yes, and that is largely the point of the formula. Write a single typical term carrying one unknown counter, collect the variable into one exponent, then solve the small equation that makes the exponent match the power you want. Evaluate that one term alone, and read a fractional or out of range solution as proof that no such term exists.
Exam move
Rehearse the counter move in both directions in one sitting: take three sums that begin at 1 and rewrite them to begin at 0, then three that begin at 5 and rewrite them to begin at 1, checking the first term, the last term and the term count on every one.
Follow that with six brackets raised to a power, writing only the general term with its exponent collected and stopping before any coefficient is worked out, until computing last has become the habit.
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