Chapter 2 · Sequences · Concept 3
2-15 Sum of (Arithmetic) × (Geometric)
3 × 1 + 5 × 3 + 7 × 9 + 9 × 27 + … . The numbers in front go up in twos, and the ones behind multiply by three. Neither kind of sum formula touches a mixture like this — but the trick from lesson 2-9 still works, with one difference at the end.
Guiding question
What happens when the two kinds of sequence are multiplied together?
Learning outcome. Find the sum of the product of an arithmetic sequence and a geometric sequence.
Work out S − rS
For the sum S of (an arithmetic sequence) × (a geometric sequence with ratio r), write S, then write rS underneath it lined up by powers, and subtract.
use S − rS
What is different from lesson 2-9. There, the middle cancelled to nothing. Here the numbers in front are not equal — they are one step of the arithmetic sequence apart — so each column leaves that step behind. The middle becomes a geometric series, and you finish it with the 2-9 formula.
The mistake the book warns about by name:
When you write rS underneath, its last term sticks out past the end of the top row with nothing above it. Subtracting brings it down with a minus sign. Forgetting that minus is the single commonest error on these two pages.
| S = | 3×3⁰ | 5×3¹ | 7×3² | 9×3³ | 11×3⁴ |
Find S = 3×1 + 5×3 + 7×32 + 9×33 + … + (2n + 1)×3n−1.
- 1
Name the two sequences before doing anything.
arithmetic: 3, 5, 7, … (step 2) · geometric: 1, 3, 9, … (ratio 3)The ratio tells you what to multiply by. Here it is 3.
Work with a partner. One of you writes S, the other writes rS directly underneath — on the same lines, shifted one place. Then subtract together, out loud, column by column.
5, 5 + 9, 5 + 9 + 13, 5 + 9 + 13 + 17, … + (4k + 1)
Find the k-th term of this sequence, and then the sum of its first n terms.
In lesson 2-9 the shift trick wiped the middle out completely. Here it leaves a geometric series behind. What is it about the numbers in front that makes the difference?
Compare the coefficients in each column.
Losing the minus on the last term.
The last term of rS has nothing above it, so it comes down subtracted. The textbook puts this in a red box for a reason.
Misaligning the two rows.
rS must be written one place to the RIGHT, so equal powers sit in the same column. Draw the columns if it helps.
Forgetting the middle needs its own formula.
What survives is a geometric series, not a finished answer. Sum it with a(rⁿ − 1)/(r − 1).
Sign slip when dividing by (1 − r).
With r = 3 you are dividing by −2. Two negatives make a positive — check your last sign against a small case.
Not testing at n = 1.
Every answer here can be checked in five seconds by putting n = 1 and comparing with the first term.
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