Compound interest and geometric sequences in Mathematik (CBK) at WU

The second of the nine topics in the syllabus of Mathematik in the CBK at WU Vienna is called "Elementare Finanzmathematik" (elementary financial mathematics). The syllabus does not list its subtopics; it says a detailed overview is on the maths pages on Canvas.¹ In the textbook by Birgit Rudloff, who leads the course, and Achim Zeileis, the second chapter has the same title. Its first three sections are about how capital grows with interest, how long it takes to reach a target, and how to add up amounts that grow by the same factor from one period to the next.

You need powers, roots and, for the time to a target, the logarithm, all on the calculator. According to the syllabus, help is available in the forum on Canvas and at the walk-in tutorial, where you work through exercises with a tutor on hand.¹ Four mistakes are typical: simple interest where compound interest is meant, a number of years rounded the wrong way, the interest rate put where the factor one plus the rate belongs, and a miscounted number of terms. German terms are given in brackets.

Simple and compound interest: Mathematik (CBK)

Capital $K_0$ is invested at an interest rate $i$ per year; at 6% that is $i=0.06$. With simple interest (einfache Verzinsung) interest is paid on the starting capital only, and after $n$ years the capital is $K_n=K_0\cdot (1+n\cdot i)$. With compound interest (Zinseszins, zusammengesetzte Verzinsung) the interest is added to the capital at the end of each year and earns interest itself from then on. So each year the capital is multiplied by the same factor $q=1+i$, the accumulation factor (Aufzinsungsfaktor):

$$K_n=K_0\cdot q^n$$

With simple interest every year brings the same amount, 6% of 300 MU (monetary units), which is 18 MU, and after 10 years you have 480 MU. With compound interest, each year also brings interest on the interest so far: in the second year an extra 6% of 18 MU, which is 1.08 MU, so 19.08 MU in all. In the third year the 18 MU are joined by 6% of the 18 + 19.08 = 37.08 MU of interest earned so far, which is 2.22 MU, and so each year's gain is larger than the last. After 10 years you have $300\cdot 1.06^{10}\approx 537.25$ MU, which is 57.25 MU more than with simple interest.

Our calculation

Numbers from the article’s example (our own numbers, not an exam question): capital 300 MU, interest rate 6% a year, 10 years.

Capital after n years with compound interest: K_n = 300 · 1.06ⁿ. Interest in year n: K_n − K_n−1 = 0.06 · K_n−1 = 18 + 0.06 · (K_n−1 − 300), that is 18 MU on the 300 MU and 6% on all the interest so far. Simple interest: 18 · n.

year	capital at year end	interest in the year	of which interest on interest	interest in total	simple interest in total
0	300.00			0.00	0.00
1	318.00	18.00	0.00	18.00	18.00
2	337.08	19.08	1.08	37.08	36.00
3	357.30	20.22	2.22	57.30	54.00
4	378.74	21.44	3.44	78.74	72.00
5	401.47	22.72	4.72	101.47	90.00
6	425.56	24.09	6.09	125.56	108.00
7	451.09	25.53	7.53	151.09	126.00
8	478.15	27.07	9.07	178.15	144.00
9	506.84	28.69	10.69	206.84	162.00
10	537.25	30.41	12.41	237.25	180.00

After 10 years: 300 · 1.06¹⁰ ≈ 537.25 MU, interest in total 237.25 MU instead of 180 MU, a difference of 57.25 MU. Values rounded to the cent.

The formula can be solved for any of its quantities. Solved for $K_0$ it gives the present value (Barwert), the amount you must invest today to have $K_n$ after $n$ years: $K_0=K_n/q^n$, which is called discounting. To have 480 MU in 10 years at 6%, you need $480/1.06^{10}\approx 268.03$ MU today. Solved for $q$ it gives the rate that turns 300 MU into 480 MU in 10 years, through the tenth root: $q=(480/300{)}^{1/10}=1.6^{0.1}$, on the calculator $1.6$ to the power of $0.1$, so a rate of about 4.81%.

Working with $K_0\cdot (1+n\cdot i)$ where compound interest is meant gives simple interest, 57.25 MU too little in the example. Careful reading is what helps here: if the question says interest is added to the capital or earns interest itself, it is compound interest. Taking $q=0.06$ instead of $1.06$ gives a result below the starting capital, so that mistake shows at once.

Time to reach a target with the logarithm: Mathematik (CBK)

When the number of years (Laufzeit) is the unknown, $n$ sits in the exponent. The question of when 300 MU at 6% grow to at least 480 MU is the equation $300\cdot 1.06^n=480$, so $1.06^n=1.6$. The logarithm brings the exponent down, because for any logarithm $\ln (q^n)=n\cdot \ln q$. Take logarithms of both sides, $n\cdot \ln 1.06=\ln 1.6$, and divide by $\ln 1.06$:

$$n=\frac{\ln (K_n/K_0)}{\ln q}=\frac{\ln 1.6}{\ln 1.06}\approx 8.07$$

It does not matter which logarithm you use as long as it is the same above and below the line; on the calculator you use ln.

After 8 years the capital is only $300\cdot 1.06^8\approx 478.15$ MU, just short of the target, after 9 years $506.84$ MU. If interest is credited at the end of each year, or the question asks for full years, the target is reached only after 9 years: round up, even when the decimal part is as small as here. The same two calculations check any answer of this kind: with the rounded number the capital must reach the target, with one year fewer it must fall short.

With simple interest it would take 10 years, because only 18 MU are added each year. A number of full years that is longer with compound interest than with simple interest is therefore always wrong. How long capital takes to double depends only on the rate, not on the amount: at 6%, $n=\ln 2/\ln 1.06\approx 11.90$, so 12 full years.

Geometric sequences and their sums: Mathematik (CBK)

The capital values year by year form a geometric sequence (geometrische Folge): each term is the one before times the same factor $q$. In arithmetic sequences the same amount is added each time; here you multiply by the same factor. With first term $a_1$, the $n$-th term is $a_n=a_1\cdot q^{n-1}$, because the factor has been applied $n-1$ times by then. The capital formula has $q^n$ because that sequence starts with $K_0$, the capital after zero years: after $n$ years the factor has been applied $n$ times.

An app gains 32 new users in its first month and after that 50% more each month than the month before: 32, 48, 72, 108, 162 and 243. To find their sum $s_6$, multiply the whole sum by $q=1.5$. Each term turns into the next one, 32 into 48, 48 into 72 and so on, and the last term 243 turns into $243\cdot 1.5=364.5$, the seventh term. Subtract the original sum and five terms cancel, leaving only this new last term and the old first one: $1.5\cdot s_6-s_6=364.5-32=332.5$. The left side is $0.5\cdot s_6$, so $s_6=332.5/0.5=665$.

s₆ = 332.5 / 0.5 = 665

Our calculation

Numbers from the article’s example (our own numbers, not an exam question): a₁ = 32, q = 1.5, n = 6.

The terms: 32 · 1.5 = 48, 48 · 1.5 = 72, 72 · 1.5 = 108, 108 · 1.5 = 162, 162 · 1.5 = 243. Multiplied by 1.5, each term turns into the next one, and the last one turns into 243 · 1.5 = 364.5, the seventh term a₇.

Upper row minus lower row: 48, 72, 108, 162 and 243 are in both and cancel. What remains is 364.5 − 32 = 332.5. The left side is 1.5 · s₆ − s₆ = 0.5 · s₆, so s₆ = 332.5 / 0.5 = 665.

Check, added one by one: 32 + 48 + 72 + 108 + 162 + 243 = 665. With the sum formula: 32 · (1.5⁶ − 1) / (1.5 − 1) = 32 · 10.390625 / 0.5 = 665.

The same step gives the sum of the first $n$ terms in general, for any factor $q\ne 1$:

$$s_n=a_1\cdot \frac{q^n-1}{q-1}$$

The exponent is the number of terms. Using $q^5$ for six months, because the sixth term is $a_1\cdot q^5$, adds up only five terms. Here too $q$ is the factor: for growth of 5% it is $1.05$.

Practice questions on interest and geometric sequences: Mathematik (CBK)

The exam has 20 multiple-choice questions with 5 options each, exactly one of them right. A right answer earns 1 point. Your first 3 wrong answers cost nothing, every further one costs 0.25 points, and you pass with 11 points.¹

We wrote the three questions for this article in that format. The scenarios, numbers and wording are our own; none of them comes from a WU exam or from the textbook. Every wrong option is the result of a typical mistake. Work each one out first; the solution then shows the working, the typical mistake and how to rule options out.

Practice question 1Elementary financial mathematics

How much must you invest today at 4% p.a. with compound interest to have 3,000 MU (monetary units) in 5 years (rounded)?

1. a)2,400.00 MU
2. b)2,446.12 MU
3. c)2,465.78 MU
4. d)2,500.00 MU
5. e)2,564.41 MU

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. This is the present value: K₅ = K₀ · 1.04⁵ = 3,000, so K₀ = 3,000 / 1.04⁵.
2. 1.04⁵ ≈ 1.2166529.
3. K₀ = 3,000 / 1.2166529 ≈ 2,465.78 MU.

a)

Wrong. Simple interest taken off the final amount: 3,000 · (1 − 5 · 0.04).

0 or −0.25

b)

Wrong. Discounted with 1 − i, that is multiplied by 0.96⁵ instead of divided by 1.04⁵.

0 or −0.25

c)

Right. Present value K₀ = 3,000 / 1.04⁵ ≈ 2,465.78.

+1

d)

Wrong. Simple interest instead of compound interest: 3,000 / (1 + 5 · 0.04).

0 or −0.25

e)

Wrong. Discounted for one year too few: 3,000 / 1.04⁴.

0 or −0.25

–

Left blank.

0

Typical mistake. Simple interest instead of compound interest. Dividing by 1 + 5 · 0.04 = 1.2 leaves out the interest on interest and gives 2,500 MU. With compound interest you divide by 1.04⁵, which is more than 1.2.

Ruling out before you finish. Compound interest grows money faster than simple interest, so today you need less than with simple interest, less than 3,000 / 1.2 = 2,500 MU. And dividing by 1.04 takes off less than 4% each year, so over five years less than 5 · 4% = 20% of 3,000 MU: the present value is above 2,400 MU. That rules out options a, d and e and leaves two.

A guess among the 2 left has a 1-in-2 chance. It is worth 0.5 points on average while your three free wrong answers last, and 0.375 after that.

Practice question 2Elementary financial mathematics

Capital of 2,000 MU (monetary units) is invested at 4.5% p.a.; interest is credited at the end of each year and earns interest from then on. After how many full years is it at least 3,400 MU for the first time?

1. a)12
2. b)13
3. c)15
4. d)16
5. e)28

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. 2,000 · 1.045ⁿ = 3,400, so 1.045ⁿ = 1.7.
2. Take logarithms: n = ln 1.7 / ln 1.045 ≈ 0.5306 / 0.0440 ≈ 12.06.
3. After 12 years it is 3,391.76 MU, after 13 years 3,544.39 MU. The answer is 13.

a)

Wrong. Correct calculation, rounded down: after 12 years it is only 3,391.76 MU.

0 or −0.25

b)

Right. n = ln 1.7 / ln 1.045 ≈ 12.06, rounded up: after 13 years it is 3,544.39 MU.

+1

c)

Wrong. Worked with simple interest, 90 MU every year, and rounded down: 1,400 / 90 ≈ 15.6.

0 or −0.25

d)

Wrong. Worked with simple interest: 1,400 / 90 ≈ 15.6, rounded up.

0 or −0.25

e)

Wrong. Two different logarithms mixed, ln on top and log base 10 below: ln 1.7 / log 1.045 ≈ 27.76, rounded up.

0 or −0.25

–

Left blank.

0

Typical mistake. Rounding down. n ≈ 12.06 looks like 12, but after 12 years about 8 MU are still missing. Interest only arrives at the end of a year, so the target is reached only after 13 full years.

Ruling out before you finish. Try a number; one power is enough: 1.045¹⁴ ≈ 1.85, which is more than 3,400 / 2,000 = 1.7. So after 14 years the target has certainly been reached, and the answer is at most 14. That rules out options c, d and e and leaves two.

A guess among the 2 left has a 1-in-2 chance. It is worth 0.5 points on average while your three free wrong answers last, and 0.375 after that.

Practice question 3Elementary financial mathematics

A licence fee is 2,000 MU (monetary units) in the first year and rises by 5% every year after that. How much is paid in total over the first 10 years (without interest, rounded)?

1. a)20,000.00 MU
2. b)22,053.13 MU
3. c)24,500.00 MU
4. d)25,155.79 MU
5. e)26,413.57 MU

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. The fees are a geometric sequence with a₁ = 2,000 and q = 1.05; the question asks for the sum of 10 terms.
2. 1.05¹⁰ ≈ 1.62889463.
3. s₁₀ = 2,000 · (1.62889463 − 1) / 0.05 ≈ 25,155.79 MU.

a)

Wrong. The increase forgotten: 10 · 2,000.

0 or −0.25

b)

Wrong. Only 9 terms counted, because the 10th fee is 2,000 · 1.05⁹: 2,000 · (1.05⁹ − 1) / 0.05.

0 or −0.25

c)

Wrong. The increase taken as a fixed amount, 5% of 2,000 = 100 MU more each year: an arithmetic instead of a geometric sequence, 10 · (2,000 + 2,900) / 2.

0 or −0.25

d)

Right. s₁₀ = 2,000 · (1.05¹⁰ − 1) / 0.05 ≈ 25,155.79.

+1

e)

Wrong. The increase applied from the first year already, so with 2,000 · 1.05 = 2,100 as the first term: 2,100 · (1.05¹⁰ − 1) / 0.05.

0 or −0.25

–

Left blank.

0

Typical mistake. The wrong number of terms. The exponent in the sum formula is how many terms are added, here 10. The 10th fee is 2,000 · 1.05⁹; carrying that 9 into the sum formula adds only 9 years and gives 22,053.13 MU.

Ruling out before you finish. If the fee rises by 5% of the previous year’s fee, it rises by at least 100 MU every year, and by more from the third year on. So the total is above the total with a fixed rise of 100 MU, 10 · (2,000 + 2,900) / 2 = 24,500 MU. That rules out options a, b and c and leaves two.

A guess among the 2 left has a 1-in-2 chance. It is worth 0.5 points on average while your three free wrong answers last, and 0.375 after that.

Calculator, formulas and time in Mathematik (CBK)

WU does not publish which of the 20 questions covers which chapter. This article follows the structure of the textbook, where interest models, durations and geometric sequences (Verzinsungsmodelle, Laufzeiten, Geometrische Folgen) are the first three sections of the second chapter.²

You may use a calculator, but not one with functions for differential calculus, integration or matrices, not one that solves linear systems, and not one with a text memory.¹ For this chapter you need powers, roots and the logarithm. Check before the exam that your model has an ln key and that you can enter a power with a decimal exponent.

WU's guideline for conducting exams bans formula sheets and watches of any kind unless the syllabus says otherwise.³ So you cannot bring your own formula sheet, and WU does not say whether one is handed out. Know the formulas for $K_n$ and $s_n$ by heart. The number of years needs no formula of its own: take logarithms of $K_n=K_0\cdot q^n$, as shown above.

The exam plan reserves two hours for the whole exam; the syllabus does not say how long you actually write.⁴ Spread over 20 questions, that is at most six minutes each.

Next step after interest and geometric sequences: Mathematik (CBK)

1. Do the three questions above before opening the solutions. Write down $K_n=K_0\cdot q^n$ and the sum formula from memory first.
2. If you got one wrong, find the mistake in the solution that matches your answer. If it was the number of years, from now on check with the rounded number and with one year fewer. If it was a sum, count the terms before you calculate. For more practice, see the second chapter, Elementare Finanzmathematik, of the free textbook by Birgit Rudloff and Achim Zeileis.⁵
3. Annuities, that is regular equal payments, are sums of exactly this kind; they and interest compounded several times a year or continuously are covered in our article on annuities and continuous compounding. Exam dates and registration are in the overview of the Mathematik (CBK) exam.

Sources

1. 1Syllabus 0003 Mathematik (LVP), Wintersemester 2026/27WU Wien, Vorlesungsverzeichnis · undated page · checked on 27 September 2026
2. 2Rudloff, Zeileis: Mathematik für Wirtschaftswissenschaften, Kapitel 2: Elementare Finanzmathematikmathe4wiwi.org · undated page · checked on 27 September 2026
3. 3Richtlinie zur Abhaltung von Präsenzprüfungen, 2026 (PDF)WU Wien · document dated 2 March 2026 · checked on 27 September 2026
4. 4Prüfungsplan Semestermitte WiSe 2026/27, Prüfungswoche November 2026 (PDF)WU Wien, Prüfungsorganisation · document dated 30 March 2026 · checked on 27 September 2026
5. 5Rudloff, Zeileis: Mathematik für Wirtschaftswissenschaften (Online-Buch, Version 2023)mathe4wiwi.org · undated page · checked on 26 September 2026