Sigma notation and arithmetic series

About this lesson

Precalculus › Sequences and series

To save a PDF, choose “Save as PDF” in the print dialog.

Sigma notation and arithmetic series

Sequences and series. sequences, series, arithmetic sequence, geometric sequence, sigma notation, geometric series, binomial theorem, Pascal's triangle

Warm-up

Three quick questions from earlier lessons.

Discover

To add 1+2+⋯+1001 + 2 + \dots + 100, pair the first and last terms: 1+1001 + 100, 2+992 + 99, … How many pairs, and what is each pair worth?

  1. 50 pairs of 101
  2. 100 pairs of 101

Key idea

∑k=pqak\displaystyle\sum_{k=p}^{q} a_k means ap+ap+1+⋯+aqa_p + a_{p+1} + \dots + a_q: it has q−p+1q - p + 1 terms.

The sum of the nn terms of an arithmetic sequence is Sn=n(a1+an)2S_n = \dfrac{n(a_1 + a_n)}{2}: the number of terms times the average of the first and last. In particular 1+2+⋯+n=n(n+1)21 + 2 + \dots + n = \dfrac{n(n + 1)}{2}.

Worked example

Find ∑k=120(3k+2)\displaystyle\sum_{k=1}^{20} (3k + 2).

First term 55, last term 6262
k = 1 and k = 20.
S=20(5+62)2S = \dfrac{20(5 + 62)}{2}
20 terms.
S=670S = 670
10 · 67.

Your turn, guided

Your turn: 2+4+6+⋯+402 + 4 + 6 + \dots + 40.

n=20n = 20
40 ÷ 2: 20 terms.
S=420S = 420
20 · 42 / 2.

Practice ★

  1. How many terms does ∑k=317(2k+1)\displaystyle\sum_{k=3}^{ 17 } (2k + 1) have?
  2. How many terms does ∑k=321(2k+1)\displaystyle\sum_{k=3}^{ 21 } (2k + 1) have?
  3. How many terms does ∑k=238(2k+1)\displaystyle\sum_{k=2}^{ 38 } (2k + 1) have?

Watch out!

Practice ★★

  1. Find ∑k=130(3k+7)\displaystyle\sum_{k=1}^{ 30 } (3k + 7).
  2. Find ∑k=111(6k+8)\displaystyle\sum_{k=1}^{ 11 } (6k + 8).
  3. Find ∑k=121(6k+6)\displaystyle\sum_{k=1}^{ 21 } (6k + 6).
  1. Find 1+2+3+⋯+941 + 2 + 3 + \dots + 94.
  2. Find 1+2+3+⋯+801 + 2 + 3 + \dots + 80.

Exit ticket

Three questions to check you have got it.

  1. Find ∑k=130(3k+7)\displaystyle\sum_{k=1}^{ 30 } (3k + 7).
  2. Find ∑k=111(6k+8)\displaystyle\sum_{k=1}^{ 11 } (6k + 8).
  3. Find ∑k=121(6k+6)\displaystyle\sum_{k=1}^{ 21 } (6k + 6).

Summary card

Arithmetic series

  • ∑k=pq\displaystyle\sum_{k=p}^{q} has q−p+1q - p + 1 terms.
  • Sn=n(a1+an)2S_n = \dfrac{n(a_1 + a_n)}{2}.
  • 1+2+⋯+n=n(n+1)21 + 2 + \dots + n = \dfrac{n(n + 1)}{2}.

Watch out: count the terms with +1+1.

CC BY-SA 4.0