Chapter 7 · Binomial Theorem

7.2 Applications of the Binomial Expansion

Split a number into a round part and a small part, and the binomial theorem does the arithmetic: $99^5$ exactly, $(1.01)^{10}$ to four decimal places, which of two huge powers is bigger, and why $9^n-8n-1$ is always a multiple of $64$.

What is $99^5$? A calculator with an eight-digit display cannot tell you: the answer has ten digits. Multiplying $99$ by itself five times by hand is slow. But $99$ is just one less than $100$, and powers of $100$ are easy — they are a $1$ followed by zeros. Writing $99^5=(100-1)^5$ and expanding with the binomial theorem of Section 7.1 turns the problem into adding six simple numbers. The same idea, "a round number plus a small correction", is how engineers and scientists estimate quantities in their heads.

In this section we put the binomial theorem to work in three ways. First, we evaluate powers exactly, such as $99^5$ or $102^4$. Second, we approximate powers such as $(1.01)^{10}$ or $(0.98)^5$, where the terms of the expansion of $(1+x)^n$ shrink so quickly that the first few give the answer to several decimal places — and we use the same expansion to decide which of two large numbers, such as $(1.01)^{1000}$ and $1000$, is bigger. Third, we prove divisibility results and find remainders: every term of $(1+x)^n$ except the first two contains $x^2$. Banks use these expansions for compound interest, and demographers for population growth.

Evaluating Powers Exactly

To evaluate $N^n$ for a whole number $N$ close to a power of $10$, write $N=10^k\pm d$ with a small $d$, expand with Theorem 7.1 (or Theorem 7.2 for the minus sign), and add. Because the powers of $10^k$ are just shifted digits, each term is easy, and only the small number $d$ has to be raised to powers.

🧭 Problem-solving strategy — Evaluating or estimating a power
  1. Split the base as (easy number) $\pm$ (small number): $99=100-1$, $\ 10.2=10+0.2$, $\ 0.98=1-0.02$.
  2. Expand with the binomial theorem; for $(a-b)^n$ the signs alternate.
  3. Work out each term separately. For an exact answer keep every term; for an estimate stop when the terms become smaller than the accuracy you need.
  4. Add the terms, lining up the place values carefully.
  5. Check the last digit, or the size, by a quick independent argument.
✏️ Example 1 — Evaluating $99^5$
Use the binomial theorem to evaluate $99^5$.
Split $99^5=(100-1)^5$, with row $5$: $1,5,10,10,5,1$ and alternating signs.
Terms $100^5-5(100)^4+10(100)^3-10(100)^2+5(100)-1$ $=10\,000\,000\,000-500\,000\,000+10\,000\,000$ $-100\,000+500-1$.
Add Group the positive and negative terms: $10\,010\,000\,500-500\,100\,001$.
Answer $99^5=9\,509\,900\,499$.
Check The last digit: $9^5$ ends in $9$ (odd powers of $9$ end in $9$) ✓. Size: $99^5\lt100^5=10^{10}$ ✓.
✏️ Example 2 — A decimal power exactly
Evaluate $(10.2)^4$ exactly.
Split $(10.2)^4=(10+0.2)^4$, coefficients $1,4,6,4,1$.
Terms $10^4+4(10)^3(0.2)+6(10)^2(0.2)^2+4(10)(0.2)^3+(0.2)^4$ $=10000+800+24+0.32+0.0016$.
Answer $(10.2)^4=10824.3216$.
Check $(10.2)^2=104.04$ and $104.04^2=10824.3216$ ✓.
Interactive 1 — Expansion Lab: One Term at a Time

Pick a power. Each row of the picture is one term ${}^nC_r\,A^{n-r}B^r$ of the expansion of $(A+B)^n$; the bar shows its size on a logarithmic scale (each grid step is a factor of $10$), blue for positive and red for negative terms. Press Play to add the terms one by one and watch the running total land exactly on the answer. The highlighted term is the one being added.

term being added—
running total—
exact value—

Approximations and Comparing Large Powers

Now take $(1+x)^n$ with $x$ small, say $x=0.01$. By Theorem 7.2, $$(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+\dfrac{n(n-1)(n-2)}{6}x^3+\cdots+x^n.$$ Each new term carries one more factor of $x$, so when $n|x|$ is small the terms shrink very fast: for $(1.01)^{10}$ they are $1$, $\ 0.1$, $\ 0.0045$, $\ 0.00012$, $\dots$. The first two terms give the quick estimate $(1+x)^n\approx1+nx$, and each further term adds roughly one or two more correct decimal places.

📐 Theorem 7.3 — A lower bound for $(1+x)^n$

Let $x\ge0$ and let $n$ be a positive integer. Then $$(1+x)^n\ge1+nx,$$ and for $n\ge2$ also $(1+x)^n\ge1+nx+\dfrac{n(n-1)}{2}x^2$. More generally, $(1+x)^n$ is at least the sum of any number of the first terms of its expansion.

Proof. By Theorem 7.2, $(1+x)^n={}^nC_0+{}^nC_1x+{}^nC_2x^2+\cdots+{}^nC_nx^n$. Every coefficient ${}^nC_r$ is positive and every power $x^r$ is $\ge0$ because $x\ge0$, so every term is $\ge0$. Leaving out some of the terms can therefore only make the sum smaller (or keep it equal), which is the claim. $\blacksquare$

⚠️ Know when to stop — and when not to

The estimate $1+nx$ is good only when $nx$ is small. For $(1.01)^{10}$, $nx=0.1$ and $1+nx=1.1$ is close to the true $1.1046\ldots$. For $(1.01)^{1000}$, $nx=10$ and $1+nx=11$ — but the true value is about $20959$! When $nx$ is not small, the later terms are larger, not smaller, and stopping early gives only a lower bound (Theorem 7.3), not an approximation.

✏️ Example 3 — Four decimal places from three or four terms
Find, correct to four decimal places, (a) $(1.01)^{10}$ (b) $(0.98)^5$.
(a) Terms $(1+0.01)^{10}=1+10(0.01)+45(0.01)^2+120(0.01)^3+210(0.01)^4+\cdots$ $=1+0.1+0.0045+0.00012+0.0000021+\cdots$.
(a) Answer The terms after the fourth add less than $0.000003$, which cannot change the fourth decimal place: $1.10462\ldots\approx1.1046$.
(b) Terms $(1-0.02)^5=1-5(0.02)+10(0.02)^2-10(0.02)^3+5(0.02)^4-(0.02)^5$ $=1-0.1+0.004-0.00008+0.0000008-\cdots$.
(b) Answer $1-0.1+0.004-0.00008=0.90392$, and the remaining terms change it by less than $0.000001$: $(0.98)^5\approx0.9039$. (The exact value is $0.9039207968$.)
✏️ Example 4 — Which is larger?
(a) Which is larger, $(1.02)^{500}$ or $10$? (b) Which is larger, $(1.01)^{1000}$ or $1000$?
(a) By Theorem 7.3 with $x=0.02$, $n=500$: $(1.02)^{500}\ge1+500(0.02)=11\gt10$. So $(1.02)^{500}$ is larger.
(b) Two terms fail $1+1000(0.01)=11$ is far below $1000$, so two terms prove nothing here. Keep adding terms: ${}^{1000}C_2(0.01)^2=\dfrac{1000\times999}{2}\times10^{-4}=49.95$, and so on.
(b) More terms The first six terms are about $1$, $10$, $49.95$, $166.17$, $414.17$ and $825.03$; their sum is about $1466$. By Theorem 7.3, $(1.01)^{1000}$ is at least this sum.
(b) Answer $(1.01)^{1000}\gt1466\gt1000$, so $(1.01)^{1000}$ is larger. (A calculator gives about $20959$: a $1\%$ increase repeated $1000$ times multiplies a quantity more than twenty thousand times.)
Interactive 2 — Approximation Lab

The dots are the sums of the first $1, 2, 3, \dots$ terms of the expansion of $(1+x)^n$; the dashed green line is the exact value. Press Play to step $n$ from $1$ to $40$. For small $nx$ the second dot already sits almost on the line and the error shrinks by a large factor with every extra term. Make $x$ larger and watch how many terms you need. A negative $x$ makes the dots zig-zag above and below the line.

0.0110
$1+nx$—
three terms—
exact $(1+x)^n$—
error of $1+nx$—

Divisibility and Remainders

Write $(1+x)^n=1+nx+\left[{}^nC_2x^2+{}^nC_3x^3+\cdots+x^n\right]$. Every term inside the square bracket has a factor $x^2$, and the coefficients ${}^nC_r$ are whole numbers. When $x$ is an integer, this gives a powerful divisibility test.

📐 Theorem 7.4 — Divisibility from the binomial theorem

Let $x$ be an integer and $n$ a positive integer. Then

  • $(1+x)^n-1$ is divisible by $x$; in other words, $(1+x)^n$ leaves remainder $1$ on division by $x$ (for $x\ge2$);
  • $(1+x)^n-1-nx$ is divisible by $x^2$.

Proof. By Theorem 7.2, $(1+x)^n-1-nx$ $={}^nC_2x^2+{}^nC_3x^3+\cdots+{}^nC_nx^n$ $=x^2\left({}^nC_2+{}^nC_3x+\cdots+{}^nC_nx^{n-2}\right)$ (for $n=1$ both sides are $0$). The bracket is an integer, since the ${}^nC_r$ and $x$ are integers, so the right side is a multiple of $x^2$. Similarly $(1+x)^n-1=x\left(n+{}^nC_2x+\cdots+x^{n-1}\right)$ is a multiple of $x$. $\blacksquare$

✏️ Example 5 — A multiple of $64$
Prove that $9^n-8n-1$ is divisible by $64$ for every positive integer $n$.
Split $9=1+8$, so $9^n=(1+8)^n=1+8n+{}^nC_2\,8^2+{}^nC_3\,8^3+\cdots+8^n$.
Rearrange $9^n-8n-1=8^2\left({}^nC_2+{}^nC_3\cdot8+\cdots+8^{n-2}\right)=64k$, where $k$ is an integer.
Conclude So $64$ divides $9^n-8n-1$ — this is Theorem 7.4 with $x=8$. $\blacksquare$
Check $n=1$: $0$ ✓; $\ n=2$: $81-17=64$ ✓; $\ n=3$: $729-25=704=64\times11$ ✓.
✏️ Example 6 — Remainders without long division
Find the remainder when (a) $8^{30}$ is divided by $7$ (b) $6^{25}$ is divided by $7$ (c) find the last two digits of $11^{10}$.
(a) $8^{30}=(1+7)^{30}=1+7\left(30+{}^{30}C_2\cdot7+\cdots\right)$: a multiple of $7$ plus $1$. Remainder $1$.
(b) Set up $6^{25}=(7-1)^{25}$. Every term except the last contains a factor $7$, and the last term is $(-1)^{25}=-1$. So $6^{25}=7m-1$ for some whole number $m$.
(b) Answer A remainder must be between $0$ and $6$: $7m-1=7(m-1)+6$, so the remainder is $6$. (Students often write "$-1$" — that is not a remainder.)
(c) $11^{10}=(1+10)^{10}=1+10\cdot10+{}^{10}C_2\cdot10^2+\cdots$. Every term from the third on is a multiple of $100$, so the last two digits are those of $1+100$: they are $01$. (Indeed $11^{10}=25\,937\,424\,601$ ✓.)
⚠️ Common mistakes in applications

1. Dropping terms in an "exact" calculation. $99^5$ needs all six terms; leaving out the $-1$ at the end gives a wrong last digit.

2. Signs with $(a-b)^n$. In $(100-1)^5$ the terms alternate $+,-,+,-,+,-$; the last term is $(-1)^5=-1$.

3. Using $1+nx$ when $nx$ is not small. It is then only a lower bound (for $x\ge0$), not an estimate — see Example 4(b).

4. Place values with decimals. $(0.02)^3=0.000008$, not $0.0008$: count the decimal places, $2\times3=6$.

5. Negative "remainders". If the expansion gives $7m-1$, rewrite it as $7(m-1)+6$: the remainder is $6$.

📘 In the CBSE syllabus — school assessment only (not in the year-end board exam)

The general and middle terms are taught and tested in school, but not in the year-end board exam; JEE Main and JEE Advanced both list them and regularly ask for the middle term(s) of an expansion and for the sum of its coefficients.

Middle terms. $(a+b)^n$ has $n+1$ terms. If $n$ is even there is one middle term, $T_{\frac n2+1}$; if $n$ is odd there are two, $T_{\frac{n+1}2}$ and $T_{\frac{n+3}2}$. The largest binomial coefficient ${}^nC_r$ always sits in the middle of the row. Sum of coefficients. The sum of all the coefficients of a polynomial $p(x)$ is $p(1)$; for example, the coefficients of $(3x-2)^7$ add up to $(3-2)^7=1$.

Worked example. Find the middle term of $\left(2x+\dfrac1x\right)^8$.
Solution. $n=8$ is even, so there are $9$ terms and the middle one is $T_5$, with $r=4$: $T_5={}^8C_4(2x)^4\left(\dfrac1x\right)^4=70\times16=1120$. (The powers of $x$ cancel, so this middle term is a constant.)

Exercises

I
Evaluate exactly, using the binomial theorem.
  • 1 $101^4$
  • 2 $98^5$
  • 3 $999^3$
  • 4 $(9.9)^3$
  • 5 $51^4$
  • 6 $(1.1)^5$
1 $(100+1)^4=100\,000\,000+4\,000\,000+60\,000+400+1=104\,060\,401$.
2 $(100-2)^5=10^{10}-5(10^8)(2)+10(10^6)(4)-10(10^4)(8)+5(100)(16)-32$ $=10\,000\,000\,000-1\,000\,000\,000+40\,000\,000-800\,000+8\,000-32=9\,039\,207\,968$.
3 $(1000-1)^3=1\,000\,000\,000-3\,000\,000+3\,000-1=997\,002\,999$.
4 $(10-0.1)^3=1000-30+0.3-0.001=970.299$.
5 $(50+1)^4=6\,250\,000+500\,000+15\,000+200+1=6\,765\,201$.
6 $(1+0.1)^5=1+0.5+0.1+0.01+0.0005+0.00001=1.61051$.
II
Approximations and comparisons.
  • 1 Find $(1.02)^8$ correct to four decimal places.
  • 2 Find $(0.99)^{10}$ correct to four decimal places.
  • 3 Find $(1.005)^{20}$ correct to four decimal places.
  • 4 Which is larger, $(1.1)^{10}$ or $2$? Justify without a calculator.
  • 5 Show that $(1.03)^{40}\gt2$.
  • 6 Show that $(1.001)^{2000}\gt4.99$.
  • 7 Write $2.01=2(1+0.005)$ and find $(2.01)^5$ correct to two decimal places.
Check: partial sums of $(1+x)^n$
first 4 terms—
exact (calculator)—
to 4 d.p.—
1 $1+8(0.02)+28(0.02)^2+56(0.02)^3+70(0.02)^4+\cdots$ $=1+0.16+0.0112+0.000448+0.0000112+\cdots\approx1.17166$, so $1.1717$.
2 $1-10(0.01)+45(0.01)^2-120(0.01)^3+210(0.01)^4-\cdots$ $=1-0.1+0.0045-0.00012+0.0000021\approx0.90438$, so $0.9044$.
3 $1+20(0.005)+190(0.005)^2+1140(0.005)^3+\cdots$ $=1+0.1+0.00475+0.0001425+0.000003+\cdots\approx1.10490$, so $1.1049$.
4 $(1+0.1)^{10}=1+1+45(0.01)+\cdots=2+0.45+\cdots\gt2$ (all terms positive). So $(1.1)^{10}$ is larger (it is about $2.594$).
5 Theorem 7.3: $(1.03)^{40}\ge1+40(0.03)=2.2\gt2$.
6 Three terms: $1+2000(0.001)+\dfrac{2000\times1999}{2}(0.001)^2=1+2+1.999=4.999$. By Theorem 7.3, $(1.001)^{2000}\ge4.999\gt4.99$.
7 $(2.01)^5=32(1.005)^5=32\left(1+0.025+0.00025+0.00000125+\cdots\right)\approx32\times1.02525=32.808$, so $32.81$.
III
Divisibility and remainders. Let $n$ be a positive integer.
  • 1 Prove that $11^n-10n-1$ is divisible by $100$.
  • 2 Prove that $5^n-4n-1$ is divisible by $16$.
  • 3 Find the remainder when $9^{12}$ is divided by $8$.
  • 4 Find the remainder when $5^{40}$ is divided by $6$.
  • 5 Find the remainder when $6^{31}$ is divided by $7$.
  • 6 Prove that $3^{2n+2}-8n-9$ is divisible by $64$.
  • 7 Find the last two digits of $31^4$ without multiplying it out.
1 $11^n=(1+10)^n=1+10n+10^2\left({}^nC_2+{}^nC_3\cdot10+\cdots\right)$, so $11^n-10n-1$ is $100$ times an integer (Theorem 7.4 with $x=10$).
2 $5^n=(1+4)^n=1+4n+4^2\left({}^nC_2+{}^nC_3\cdot4+\cdots\right)$, so $5^n-4n-1=16k$. Check $n=2$: $25-9=16$ ✓.
3 $9^{12}=(1+8)^{12}=1+8(\cdots)$: remainder $1$.
4 $5^{40}=(6-1)^{40}$; every term but the last has a factor $6$, and the last is $(-1)^{40}=1$. Remainder $1$.
5 $6^{31}=(7-1)^{31}=7m+(-1)^{31}=7m-1=7(m-1)+6$: remainder $6$.
6 $3^{2n+2}=9^{n+1}=(1+8)^{n+1}=1+8(n+1)+64k$. So $3^{2n+2}-8n-9=64k$. Check $n=1$: $81-17=64$ ✓.
7 $(1+30)^4=1+4(30)+6(900)+\cdots=1+120+(\text{multiples of }100)$. The last two digits are those of $121$: $21$. (Indeed $31^4=923\,521$ ✓.)
IV
Applications. Give money to the nearest paisa unless told otherwise.
  • 1 $\text{₹}\,10\,000$ is invested at $5\%$ per year, compounded yearly. Use the binomial theorem to find the amount after $4$ years.
  • 2 A town of $2\,00\,000$ people grows by $2\%$ each year. Estimate its population after $3$ years, to the nearest person.
  • 3 A phone bought for $\text{₹}\,40\,000$ loses $10\%$ of its value each year. Find its value after $3$ years.
  • 4 A sensor's reading drifts up by $0.3\%$ every hour. Use three terms of the expansion to estimate the factor by which the reading has grown after $10$ hours, and give the percentage increase.
  • 5 A machine makes ball bearings whose radius is $0.2\%$ too large. Since the volume of a sphere is proportional to $r^3$, by about what percentage is the volume too large?
1 $(1.05)^4=1+0.2+0.015+0.0005+0.00000625=1.21550625$, so the amount is $\text{₹}\,12\,155.06$.
2 $(1.02)^3=1+0.06+0.0012+0.000008=1.061208$; $\ 200\,000\times1.061208=212\,241.6$, about $2\,12\,242$ people.
3 $(0.9)^3=(1-0.1)^3=1-0.3+0.03-0.001=0.729$; value $40\,000\times0.729=\text{₹}\,29\,160$.
4 $(1.003)^{10}\approx1+10(0.003)+45(0.003)^2=1+0.03+0.000405=1.030405$: an increase of about $3.04\%$. (The exact factor is $1.030408\ldots$.)
5 $(1.002)^3=1+0.006+0.000012+0.000000008=1.006012008$: the volume is about $0.6\%$ too large — three times the error in the radius, by the $1+nx$ estimate.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: Board papers ask you to evaluate a power such as $98^4$ or $(10.1)^3$ by the binomial theorem, to compare two powers, and to prove a divisibility result of the form "$a^n-bn-1$ is divisible by $b^2$". JEE asks for remainders of large powers and for middle terms and sums of coefficients.

E1. Board · 1 mark The remainder when $8^{25}$ is divided by $7$ is (A) $0$   (B) $1$   (C) $6$   (D) $7$

Idea $8^{25}=(1+7)^{25}=1+7(\cdots)$, a multiple of $7$ plus $1$.
Trap (C) would be right for $6^{25}=(7-1)^{25}$; (D) is impossible, as a remainder on division by $7$ is less than $7$.
Answer (B) $1$.

E2. Board · 2 marks Using the binomial theorem, evaluate $98^4$.

Expand $(100-2)^4=100^4-4(100)^3(2)+6(100)^2(2)^2-4(100)(2)^3+2^4$. (1 mark)
Answer $100\,000\,000-8\,000\,000+240\,000-3\,200+16=92\,236\,816$. (1 mark)

E3. Board · 3 marks Prove that $4^n-3n-1$ is divisible by $9$ for every positive integer $n$.

Expand $4^n=(1+3)^n={}^nC_0+{}^nC_1\cdot3+{}^nC_2\cdot3^2+\cdots+{}^nC_n\cdot3^n$. (1 mark)
Rearrange $4^n-1-3n=3^2\left({}^nC_2+{}^nC_3\cdot3+\cdots+3^{n-2}\right)$ for $n\ge2$, and $4^1-3-1=0$ for $n=1$. (1 mark)
Answer The bracket is an integer, so $4^n-3n-1=9k$ is divisible by $9$. Check $n=2$: $16-7=9$ ✓; $\ n=3$: $64-10=54=9\times6$ ✓. (1 mark)

E4. Board · Case study Riya deposits $\text{₹}\,50\,000$ in a scheme that pays $1\%$ interest per month, compounded monthly, so after $12$ months her money is multiplied by $(1.01)^{12}$. (i) Estimate $(1.01)^{12}$ using the first two terms of its expansion. (1 mark) (ii) Improve the estimate using three terms, and find the corresponding amount. (1 mark) (iii) A simple-interest scheme pays $12\%$ for the year. Using four terms of the expansion, find by about how much the monthly scheme is better. (2 marks)

(i) $1+12(0.01)=1.12$. (1 mark)
(ii) $1+0.12+66(0.0001)=1.1266$; amount $50\,000\times1.1266=\text{₹}\,56\,330$. (1 mark)
(iii) Four terms Add ${}^{12}C_3(0.01)^3=220\times0.000001=0.00022$: factor $1.12682$, amount $\approx\text{₹}\,56\,341$. Simple interest gives $50\,000\times1.12=\text{₹}\,56\,000$. (1 mark)
Answer The monthly scheme is better by about $\text{₹}\,341$. (The exact factor $1.126825\ldots$ gives $\text{₹}\,56\,341.25$.) (1 mark)

E5. JEE Main The remainder when $7^{103}$ is divided by $25$ is (A) $7$   (B) $18$   (C) $24$   (D) $1$

Idea $7^2=49=50-1$, and $50$ is a multiple of $25$. So write $7^{103}=7\cdot(7^2)^{51}=7(50-1)^{51}$.
Expand Every term of $(50-1)^{51}$ except the last has a factor $50$; the last is $(-1)^{51}=-1$. So $(50-1)^{51}=25m-1$ and $7^{103}=175m-7=25(7m-1)+18$.
Trap (A) and (C) come from forgetting the factor $7$ or the sign: $-7$ is not a remainder, and $-1$ belongs to $(50-1)^{51}$, not to $7^{103}$.
Answer (B) $18$. ⏱ Fast route: $7^{103}\equiv7\cdot(-1)^{51}=-7\equiv18$ on division by $25$.

E6. JEE Advanced One or more options correct. In the expansion of $(1+x)^{20}$: (A) the sum of all the coefficients is $2^{20}$   (B) the middle term is ${}^{20}C_{10}\,x^{10}$   (C) the sum of the coefficients of the odd powers of $x$ is $2^{19}$   (D) the coefficient of $x^7$ equals the coefficient of $x^{14}$

(A) True: put $x=1$, giving $2^{20}$.
(B) True: $21$ terms, so the middle one is $T_{11}={}^{20}C_{10}x^{10}$.
(C), (D) (C) True: $x=1$ and $x=-1$ give $E+O=2^{20}$ and $E-O=0$, so $O=2^{19}$. (D) False: ${}^{20}C_7={}^{20}C_{13}$, so $x^7$ matches $x^{13}$, not $x^{14}$ (${}^{20}C_7=77520$ but ${}^{20}C_{14}=38760$).
Answer (A), (B), (C).
🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. How long does it take an investment to double? Is the bankers' "rule of 72" — doubling time $\approx\dfrac{72}{\text{rate in }\%}$ years — any good?

Why it matters. Savings schemes, loans, inflation and even population growth all compound. Money growing at rate $r$ per year (as a decimal) is multiplied by $(1+r)^n$ after $n$ years. Theorem 7.3 says $(1+r)^n\ge1+nr$, so after $n=\dfrac1r$ years the money has at least doubled — but the true doubling time is much shorter, because the later terms of the expansion add up.

What to do.

  1. For the rates $2\%$, $4\%$, $6\%$, $8\%$, $10\%$ and $12\%$, use a spreadsheet to tabulate $(1+r)^n$ for $n=1,2,\dots,50$ and find the first whole number of years $n$ with $(1+r)^n\ge2$.
  2. In the same table, write the rule-of-72 estimate $\dfrac{72}{100r}$ and the safe bound $\dfrac1r$ from Theorem 7.3.
  3. For $r=6\%$, use the first three terms $1+nr+\dfrac{n(n-1)}{2}r^2$ to find a smaller $n$ that still guarantees doubling, and compare it with the true answer.
  4. Find the current interest rate of a real savings scheme (a bank fixed deposit, a post-office scheme or a government bond) and predict its doubling time.

Tools: A spreadsheet app; a bank or post-office website for current rates

Hand in: Your table, a graph of the true doubling time against the rate, and a paragraph on how accurate the rule of 72 is and why $\dfrac1r$ is always too large.

Going further. A country's population grows at $1\%$ per year. How long does it take to double? What happens to the doubling time if the growth rate halves?

Instructor note: first whole years with $(1+r)^n\ge2$ are $36, 18, 12, 10, 8, 7$ for $r=2,4,6,8,10,12\%$ (exact doubling times $35.0, 17.7, 11.9, 9.0, 7.3, 6.1$ years); the rule of 72 gives $36, 18, 12, 9, 7.2, 6$ and $\tfrac1r$ gives $50, 25, 16.7, 12.5, 10, 8.3$. For $r=6\%$ the three-term bound first reaches $2$ at $n=13$ (true answer $12$). Population at $1\%$: about $70$ years ($69.7$); at $0.5\%$ about $139$ years.

🧪 Self-check quiz

0/5 correct

Q1. Writing $99^3=(100-1)^3$, the value of $99^3$ is

Q2. Using only the first two terms of the expansion, $(1.02)^{10}$ is approximately

Q3. The remainder when $9^{10}$ is divided by $8$ is

Q4. Which statement follows directly from $(1+x)^n\ge1+nx$ for $x\ge0$?

Q5. For every positive integer $n$, the number $5^n-4n-1$ is divisible by

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