Sequences & Series 2 — Mathslive.ie

Section 1

From a sequence to a series

A chain message goes out. On day 1, one person gets it. Each of them sends it to 2 more the next day, and so on.

Day

1

2

3

4

5

People

1

2

4

8

16

The sequence is the list of values: $1,\ 2,\ 4,\ 8,\ 16,\ \dots$

The series is what you get when you add them up:

$1 + 2 + 4 + 8 + 16 = 31$

So after 5 days, $31$ people in total have seen the message.

Notation for the sum

$T_1 + T_2 + T_3 + T_4 + T_5 = S_5$

$S_5$ means the sum of the first 5 terms.

Series sum:

$S_n = T_1 + T_2 + T_3 + \dots + T_n$

It's just the sequence with $+$ signs between the terms.

Special case: $S_1 = T_1$. With only one term, there's nothing to add to it.

Try this

For the sequence $3,\ 7,\ 11,\ 15,\ 19,\ \dots$ find $S_4$.

$S_4 = T_1 + T_2 + T_3 + T_4$

$= 3 + 7 + 11 + 15$

$S_4 = 36$

Try this

From the same sequence, what is $S_1$? What is $S_2$?

$S_1$ is just one term. $S_2$ is two terms added.

$S_1 = T_1 = 3$

$S_2 = T_1 + T_2 = 3 + 7 = 10$

$S_1 = 3,\ \ S_2 = 10$

Section 2

Sigma notation

Writing out $T_1 + T_2 + T_3 + \dots + T_n$ every time gets tiring. There's a compact way to say "sum these up":

The sigma sign $\sum$

$\displaystyle\sum_{n=\text{start}}^{\text{end}} \big(\text{rule}\big)$

Three pieces: where to start, where to end, and the rule for the term.

For example:

$\displaystyle\sum_{n=2}^{5} n^{2}\ =\ 2^{2} + 3^{2} + 4^{2} + 5^{2}$

$= 4 + 9 + 16 + 25 = 54$

Read it as: "start with $n=2$, plug in, increase $n$ by 1, plug in, $\dots$ stop when $n=5$, then add all the results."

The full series sum we wrote earlier can be written in sigma form too:

$S_n\ =\ T_1 + T_2 + T_3 + \dots + T_n\ =\ \displaystyle\sum_{k=1}^{n} T_{k}$

Try this

Evaluate $\displaystyle\sum_{n=1}^{4} (2n+1)$.

Plug in $n=1,2,3,4$, then add.

$n=1:\ 2(1)+1 = 3$

$n=2:\ 2(2)+1 = 5$

$n=3:\ 2(3)+1 = 7$

$n=4:\ 2(4)+1 = 9$

Sum $= 3+5+7+9$

$= 24$

$24$

Try this

Evaluate $\displaystyle\sum_{n=1}^{4} 2^{n}$.

Powers of 2: $2^{1}, 2^{2}, 2^{3}, 2^{4}$.

$2^{1}+2^{2}+2^{3}+2^{4}$

$= 2+4+8+16$

$= 30$

$30$

Try this

Evaluate $\displaystyle\sum_{n=2}^{5} (n^{2} - 1)$.

Start at $n=2$, end at $n=5$. Four terms in total.

$n=2:\ 4-1 = 3$

$n=3:\ 9-1 = 8$

$n=4:\ 16-1 = 15$

$n=5:\ 25-1 = 24$

Sum $= 3+8+15+24$

$= 50$

$50$

Section 3

Building $S_n$ when you know $T_n$

If you're given a rule for $T_n$, finding a series sum is straightforward: plug in, then add.

Worked example

Find $S_4$ of the series given by $T_n = 2n + 3$.

Step 1 — write out what $S_4$ means.

$S_4 = T_1 + T_2 + T_3 + T_4$

Step 2 — plug in $n = 1, 2, 3, 4$.

$T_1 = 2(1)+3 = 5$

$T_2 = 2(2)+3 = 7$

$T_3 = 2(3)+3 = 9$

$T_4 = 2(4)+3 = 11$

Step 3 — add.

$S_4 = 5 + 7 + 9 + 11$

$\boxed{S_4 = 32}$

Notice the running sums: $S_1 = 5,\ \ S_2 = 12,\ \ S_3 = 21,\ \ S_4 = 32$. Each one is just the previous sum plus the next term. That's the relationship we'll use later.

Try this

$T_n = 3n - 2$. Find $S_5$.

Get $T_1, T_2, T_3, T_4, T_5$ then add the five values.

$T_1 = 3(1)-2 = 1$

$T_2 = 3(2)-2 = 4$

$T_3 = 3(3)-2 = 7$

$T_4 = 3(4)-2 = 10$

$T_5 = 3(5)-2 = 13$

$S_5 = 1 + 4 + 7 + 10 + 13$

$S_5 = 35$

Try this

$T_n = n^{2} + 1$. Find $S_4$.

Watch the squares. $T_3$ is not $7$.

$T_1 = 1 + 1 = 2$

$T_2 = 4 + 1 = 5$

$T_3 = 9 + 1 = 10$

$T_4 = 16 + 1 = 17$

$S_4 = 2 + 5 + 10 + 17$

$S_4 = 34$

Try this

$T_n = 4n - 1$. Find $S_3$.

Only three terms. $T_1, T_2, T_3$.

$T_1 = 4(1)-1 = 3$

$T_2 = 4(2)-1 = 7$

$T_3 = 4(3)-1 = 11$

$S_3 = 3 + 7 + 11$

$S_3 = 21$

Section 4

Two consecutive sums $\rightarrow$ one term

Here's a question type that looks scary but isn't:

"Given $S_{50} = 301$ and $S_{51} = 330$, find $T_{51}$."

You don't need to find any of the individual terms. Just compare the two sums:

$S_{50} = T_1 + T_2 + \dots + T_{50}$

$S_{51} = T_1 + T_2 + \dots + T_{50} + T_{51}$

$S_{51}$ has everything $S_{50}$ has, plus one extra term: $T_{51}$. So:

$T_{51} = S_{51} - S_{50}$

$= 330 - 301$

$\boxed{T_{51} = 29}$

The same idea works for any $n$: $\ T_n = S_n - S_{n-1}$. Keep that in your back pocket — we use it in the next section.

Try this

$S_{10} = 100$ and $S_{11} = 121$. Find $T_{11}$.

$T_{11}$ is the extra term going from $S_{10}$ to $S_{11}$.

$T_{11} = S_{11} - S_{10}$

$= 121 - 100$

$T_{11} = 21$

Try this

$S_{99} = 4950$ and $S_{100} = 5050$. Find $T_{100}$.

Subtract the smaller sum from the larger.

$T_{100} = S_{100} - S_{99}$

$= 5050 - 4950$

$T_{100} = 100$

Try this

$S_{19} = 380$ and $S_{20} = 420$. Find $T_{20}$.

$T_{20} = S_{20} - S_{19}$.

$T_{20} = S_{20} - S_{19}$

$= 420 - 380$

$T_{20} = 40$

Section 5

Given $S_n$, find a formula for $T_n$

Now the more useful version. Instead of two numbers $S_{50}$ and $S_{51}$, suppose you're given a formula for $S_n$. Can you recover the formula for $T_n$?

Yes — using the same idea, written in $n$ and $n-1$.

Where the rule comes from

Stack the two sums:

$S_n\ \ \,= T_1 + T_2 + T_3 + \dots + T_{n-1} + T_n$

$S_{n-1} = T_1 + T_2 + T_3 + \dots + T_{n-1}$

Every term except $T_n$ matches up and cancels when you subtract. What's left:

The key formula:

$T_n\ =\ S_n - S_{n-1}$

Worked example

Given $S_n = n^{2} + 3n$, find a formula for $T_n$.

Step 1 — write down $S_n$.

$S_n = n^{2} + 3n$

Step 2 — get $S_{n-1}$ by replacing every $n$ with $(n-1)$.

$S_{n-1} = (n-1)^{2} + 3(n-1)$

$= n^{2} - 2n + 1 + 3n - 3$

$= n^{2} + n - 2$

Watch the minus. When subtracting $S_{n-1}$, that minus sign goes onto every term inside the bracket. This is where most marks are lost.

Step 3 — subtract.

$T_n = S_n - S_{n-1}$

$= (n^{2} + 3n) - (n^{2} + n - 2)$

$= n^{2} + 3n - n^{2} - n + 2$

$\boxed{T_n = 2n + 2}$

The free check

Always verify with $n = 1$. Recall $S_1 = T_1$, so the two formulas must agree at $n=1$.

From $T_n = 2n+2$: $T_1 = 2(1) + 2 = 4$.

From $S_n = n^{2} + 3n$: $S_1 = 1 + 3 = 4$.

Match. ✓

If the check fails, you almost certainly dropped a minus sign in Step 3. Go back and re-distribute.

Try this

$S_n = n^{2} + 2n$. Find a formula for $T_n$.

Get $S_{n-1}$ by swapping $n$ for $(n-1)$, then $T_n = S_n - S_{n-1}$.

$S_n = n^{2} + 2n$

$S_{n-1} = (n-1)^{2} + 2(n-1)$

$= n^{2} - 2n + 1 + 2n - 2$

$= n^{2} - 1$

$T_n = S_n - S_{n-1}$

$= (n^{2} + 2n) - (n^{2} - 1)$

$= n^{2} + 2n - n^{2} + 1$

Check: $T_1 = 3$, $\ S_1 = 1+2 = 3$ ✓

$T_n = 2n + 1$

Try this

$S_n = 2n^{2} + n$. Find a formula for $T_n$.

Don't forget the $2$ multiplies the whole $(n-1)^{2}$.

$S_n = 2n^{2} + n$

$S_{n-1} = 2(n-1)^{2} + (n-1)$

$= 2(n^{2} - 2n + 1) + n - 1$

$= 2n^{2} - 4n + 2 + n - 1$

$= 2n^{2} - 3n + 1$

$T_n = S_n - S_{n-1}$

$= (2n^{2} + n) - (2n^{2} - 3n + 1)$

$= 2n^{2} + n - 2n^{2} + 3n - 1$

Check: $T_1 = 3$, $\ S_1 = 2 + 1 = 3$ ✓

$T_n = 4n - 1$

Try this

$S_n = n^{2}$. Find $T_n$.

Tidy little expression. Result is a classic.

$S_n = n^{2}$

$S_{n-1} = (n-1)^{2} = n^{2} - 2n + 1$

$T_n = n^{2} - (n^{2} - 2n + 1)$

$= n^{2} - n^{2} + 2n - 1$

Check: $T_1 = 1$, $\ S_1 = 1$ ✓

$T_n = 2n - 1$

Notice that last one: the sum of the first $n$ odd numbers is $n^{2}$. $1 + 3 + 5 + 7 + 9 = 25 = 5^{2}$. Worth remembering.

Try this

$S_n = 3n^{2} - n$. Find $T_n$.

$3$ multiplies $(n-1)^{2}$. Then $-(n-1)$ is $-n+1$.

$S_n = 3n^{2} - n$

$S_{n-1} = 3(n-1)^{2} - (n-1)$

$= 3(n^{2} - 2n + 1) - n + 1$

$= 3n^{2} - 6n + 3 - n + 1$

$= 3n^{2} - 7n + 4$

$T_n = (3n^{2} - n) - (3n^{2} - 7n + 4)$

$= 3n^{2} - n - 3n^{2} + 7n - 4$

Check: $T_1 = 2$, $\ S_1 = 3 - 1 = 2$ ✓

$T_n = 6n - 4$

Try this

A student is given $S_n = n^{2} + 5n$ and writes $T_n = 2n + 5$. Use the $n=1$ check to decide whether the student is right.

Plug $n=1$ into both. If they disagree, the student is wrong.

From student's $T_n$: $T_1 = 2(1) + 5 = 7$.

From given $S_n$: $S_1 = 1 + 5 = 6$.

But $S_1$ must equal $T_1$ — and $7$ doesn't match $6$.

So the student is wrong. The correct $T_n$:

$S_{n-1} = (n-1)^{2} + 5(n-1) = n^{2} - 2n + 1 + 5n - 5 = n^{2} + 3n - 4$

$T_n = (n^{2} + 5n) - (n^{2} + 3n - 4) = 2n + 4$

Check: $T_1 = 6$, $\ S_1 = 6$ ✓

Student wrong. $T_n = 2n + 4$

Part 2 done.

You now have $S_n$ and the $T_n \leftrightarrow S_n$ machinery: build a sum, read a sigma, and recover terms from a cumulative formula.

Part 3 picks up arithmetic series (closed-form sum of an AP) and geometric series (closed-form sum of a GP) — the formulas that save you from adding 50 terms by hand.