7.1 Pascal's Triangle and the Binomial Theorem
Multiply out $(a+b)^n$ and the coefficients always come from the same triangle of numbers โ and each of them is a count ${}^nC_r$. That one counting idea proves the binomial theorem for every positive integer $n$.
You have been expanding $(a+b)^2=a^2+2ab+b^2$ since Class 8, and in Class 9 you met $(a+b)^3=a^3+3a^2b$ $+\,3ab^2+b^3$. What about $(a+b)^4$, or $(a+b)^{10}$? You could multiply $(a+b)$ by itself ten times, collecting like terms at every stage โ slow, and one slip anywhere spoils everything after it. A two-letter expression such as $a+b$, $x-2$ or $1+0.01$ is called a binomial ("two names"), and in this section we find a rule that writes down any whole-number power of a binomial in one line, with no multiplying out at all.
The rule has two faces. The first is a triangle of numbers, built by nothing more than addition, whose rows are exactly the coefficients we need โ it is called Pascal's triangle, after the French mathematician Blaise Pascal (1653), although Pingala's work on poetic metres in ancient India, the Persian mathematician al-Karaji and the Chinese mathematician Yang Hui all knew it centuries earlier. The second face is the binomial theorem, which says that the coefficients are the numbers ${}^nC_r$ of Section 6.3. We shall prove it by counting, and the counting proof will explain why the triangle works. The binomial theorem is used in probability (how likely are exactly $3$ heads in $10$ tosses?), in finance (compound interest), in quick mental estimates, and โ in Class 12 and beyond โ throughout calculus.
Powers of a Binomial and Pascal's Triangle
Let us multiply out the first few powers of $a+b$ and look only at the numbers in front of the terms:
| Power | Expansion | Coefficients |
|---|---|---|
| $(a+b)^0$ | $1$ | $1$ |
| $(a+b)^1$ | $a+b$ | $1\ \ 1$ |
| $(a+b)^2$ | $a^2+2ab+b^2$ | $1\ \ 2\ \ 1$ |
| $(a+b)^3$ | $a^3+3a^2b+3ab^2+b^3$ | $1\ \ 3\ \ 3\ \ 1$ |
| $(a+b)^4$ | $a^4+4a^3b+6a^2b^2+4ab^3+b^4$ | $1\ \ 4\ \ 6\ \ 4\ \ 1$ |
To get $(a+b)^4$ we multiplied $(a+b)^3$ by $a+b$: $$(a^3+3a^2b+3ab^2+b^3)(a+b).$$ Multiplying by $a$ gives $a^4+3a^3b+3a^2b^2+ab^3$; multiplying by $b$ gives $a^3b+3a^2b^2+3ab^3+b^4$. Adding, each new coefficient is the sum of two old ones: $1$, $\ 1+3=4$, $\ 3+3=6$, $\ 3+1=4$, $\ 1$. Four patterns stand out, and they hold for every power we try:
- $(a+b)^n$ has $n+1$ terms;
- the power of $a$ falls by one from term to term, from $a^n$ down to $a^0=1$, while the power of $b$ rises from $b^0$ to $b^n$;
- in every term the two indices add up to $n$ (for example $a^2b^2$ in $(a+b)^4$);
- the coefficients read the same forwards and backwards, start and end with $1$, and each inside coefficient is the sum of the two coefficients standing above it in the previous row.
Pascal's triangle is the triangular array of numbers built as follows. Row $0$ is the single number $1$. Every later row begins and ends with $1$, and each of its other entries is the sum of the two entries immediately above it (to the left and to the right) in the row before. Row $n$ has $n+1$ entries, and these are the coefficients in the expansion of $(a+b)^n$, read from left to right.
Why the Entries Are ${}^nC_r$
Pascal's triangle is quick for small $n$, but to reach row $20$ we would have to build all twenty rows before it. We want a formula for each entry directly. Look again at $(a+b)^3=(a+b)(a+b)(a+b)$. Multiplying out means: from each of the three brackets choose either $a$ or $b$, multiply the three choices, and add up the results for all $2\times2\times2=8$ ways of choosing. The eight products are
$aaa,\ \ aab,\ \ aba,\ \ baa,\ \ abb,\ \ bab,\ \ bba,\ \ bbb.$
The term $a^2b$ appears once for every way of deciding which one of the three brackets supplies the $b$: $aab$, $aba$, $baa$ โ that is, ${}^3C_1=3$ ways. The term $ab^2$ appears once for every choice of the two brackets that supply $b$: ${}^3C_2=3$ ways. So the coefficients $1,3,3,1$ are ${}^3C_0,{}^3C_1,{}^3C_2,{}^3C_3$. The same argument works for any power, and it explains the addition rule of Pascal's triangle too: by Theorem 6.9 of Section 6.3, ${}^{n}C_{r-1}+{}^{n}C_r={}^{n+1}C_r$ โ each entry is the sum of the two above it.
For a positive integer $n$ and $r=0,1,\dots,n$, the numbers $${}^nC_r=\dfrac{n!}{r!\,(n-r)!}$$ are called the binomial coefficients of index $n$. They are also written $\binom{n}{r}$. Entry number $r$ of row $n$ of Pascal's triangle (counting from $r=0$ at the left) is ${}^nC_r$.
The first entry of row $n$ is ${}^nC_0=1$, the second is ${}^nC_1=n$, the third is ${}^nC_2$, and so on: entry number $r$ is the $(r+1)$-th entry. In row $6$, the third entry is ${}^6C_2=15$, not ${}^6C_3=20$. Students lose marks every year by counting rows or entries from $1$ instead of $0$.
The Binomial Theorem
Let $a$ and $b$ be any real numbers (or algebraic expressions) and let $n$ be a positive integer. Then $$(a+b)^n={}^nC_0a^n+{}^nC_1a^{n-1}b+{}^nC_2a^{n-2}b^2+\cdots+{}^nC_{n-1}ab^{n-1}+{}^nC_nb^n.$$ In sigma notation, $$(a+b)^n=\sum_{r=0}^{n}{}^nC_r\,a^{n-r}b^r.$$
Proof (by counting). Write $(a+b)^n$ as the product of $n$ brackets, $(a+b)(a+b)\cdots(a+b)$. When the brackets are multiplied out, every term of the result is formed by choosing one letter, $a$ or $b$, from each bracket and multiplying the $n$ chosen letters; the expansion is the sum of these products over all possible ways of choosing.
Fix a number $r$ with $0\le r\le n$. A choice that takes $b$ from exactly $r$ brackets takes $a$ from the other $n-r$ brackets, so its product is $a^{n-r}b^r$. How many choices produce this product? A choice of this kind is completely decided by which $r$ of the $n$ brackets supply a $b$, and by Section 6.3 there are ${}^nC_r$ ways of selecting $r$ brackets out of $n$. So, after collecting like terms, $a^{n-r}b^r$ appears with coefficient exactly ${}^nC_r$. Since every choice uses some number $r$ of $b$'s between $0$ and $n$, adding over $r=0,1,\dots,n$ accounts for all the products, and $$(a+b)^n=\sum_{r=0}^{n}{}^nC_r\,a^{n-r}b^r.\ \blacksquare$$
Read the theorem together with the four patterns we spotted: there are $n+1$ terms, one for each power $b^0,b^1,\dots,b^n$; the indices of $a$ and $b$ always add up to $n$; and because ${}^nC_r={}^nC_{n-r}$ (Theorem 6.8), the coefficients are symmetric. The theorem holds for any two quantities $a$ and $b$, so we may put $a=2x$, $b=-3y$, $b=\dfrac1x$, $a=\sqrt3$ and so on โ the trick is to keep each of them in brackets while you substitute.
- Identify the first term $a$, the second term $b$ (with its sign) and the power $n$.
- Write down the coefficients ${}^nC_0,{}^nC_1,\dots,{}^nC_n$ โ from row $n$ of Pascal's triangle when $n$ is small, or from the formula.
- Write the $n+1$ terms ${}^nC_r\,(a)^{n-r}(b)^r$, keeping $a$ and $b$ in brackets; powers of a negative $b$ make the signs alternate.
- Simplify each term separately.
- Check by substituting a simple value (often $x=1$) into both the original power and your expansion.
Special Cases: $(a-b)^n$, $(1+x)^n$ and $(1-x)^n$
Three choices of $a$ and $b$ come up so often that they are worth stating on their own. Replacing $b$ by $-b$ in Theorem 7.1 changes the sign of every term with an odd power of $b$, because $(-b)^r=(-1)^rb^r$. Putting $a=1$ and $b=x$ gives the powers of $1+x$, the form used for estimates in Section 7.2.
For every positive integer $n$ and all real $a$, $b$, $x$:
- $(a-b)^n={}^nC_0a^n-{}^nC_1a^{n-1}b$ $+\,{}^nC_2a^{n-2}b^2-\cdots$ $+\,(-1)^n\,{}^nC_nb^n$;
- $(1+x)^n={}^nC_0+{}^nC_1x$ $+\,{}^nC_2x^2+\cdots+{}^nC_nx^n$, that is, $(1+x)^n=1+nx$ $+\,\dfrac{n(n-1)}{2}x^2+\cdots+x^n$;
- $(1-x)^n={}^nC_0-{}^nC_1x$ $+\,{}^nC_2x^2-\cdots$ $+\,(-1)^n\,{}^nC_nx^n$;
- ${}^nC_0+{}^nC_1+{}^nC_2+\cdots+{}^nC_n=2^n$, and ${}^nC_0-{}^nC_1+{}^nC_2-\cdots$ $+\,(-1)^n\,{}^nC_n=0$.
Proof. The first three come from Theorem 7.1 with $b$ replaced by $-b$; with $a=1$, $b=x$ (every power of $1$ is $1$); and with $a=1$, $b=-x$. For the last line put $x=1$ in the second and third results: $(1+1)^n=2^n$ and $(1-1)^n=0^n=0$ (here we need $n\ge1$). $\blacksquare$
The identity ${}^nC_0+{}^nC_1+\cdots+{}^nC_n=2^n$ is the row-sum pattern of Example 1, now proved. It also has a counting meaning: the left side counts the selections of $0$, $1$, $2$, โฆ, $n$ objects from $n$ different objects, that is, all possible selections; and making every selection is the same as deciding "in or out" for each object, which can be done in $2\times2\times\cdots\times2=2^n$ ways.
1. $(a+b)^n\ne a^n+b^n$. Already $(a+b)^2=a^2+2ab+b^2$; the cross term $2ab$ never disappears. For instance $(1+2)^3=27$, but $1^3+2^3=9$.
2. Brackets round each term. In $(2x-3y)^4$ the second term is $(2x)^2=4x^2$, not $2x^2$, and $(-3y)^3=-27y^3$, not $-3y^3$.
3. Signs with a negative second term. In $(a-b)^n$ the signs alternate $+,-,+,-,\dots$, starting with $+$. The sign of each term depends on the power of $b$, not on its position in your working.
4. Row $n$ has $n+1$ entries. $(a+b)^7$ uses row $7$, which has $8$ entries $1,7,21,35,35,21,7,1$ โ not row $6$, and not $7$ terms.
5. The theorem here needs a positive integer index. $(1+x)^{1/2}$ or $(1+x)^{-1}$ cannot be expanded this way in Class 11; their expansions go on for ever and need extra conditions on $x$.
The general term is taught and tested in school, but not in the year-end board exam; JEE Main and JEE Advanced both list it and often ask for one particular term without writing out the whole expansion.
The general term. The term containing $b^r$ in $(a+b)^n$ is the $(r+1)$-th term, written $$T_{r+1}={}^nC_r\,a^{n-r}b^r,\quad r=0,1,\dots,n.$$ To find the coefficient of a given power of $x$, or the term independent of $x$, write $T_{r+1}$ with all powers of $x$ collected, set the index of $x$ equal to the required number, and solve for $r$ โ which must be a whole number from $0$ to $n$.
Worked example. Find the term independent of $x$ in $\left(x^2+\dfrac1x\right)^9$.
Solution. $T_{r+1}={}^9C_r\,(x^2)^{9-r}\left(\dfrac1x\right)^r={}^9C_r\,x^{18-3r}$. Independent of $x$ means $18-3r=0$, so $r=6$, and the term is $T_7={}^9C_6=84$. Check: for $r=6$ the powers are $(x^2)^3\left(\tfrac1x\right)^6=x^6\cdot x^{-6}=1$ โ, and ${}^9C_6={}^9C_3=\tfrac{9\cdot8\cdot7}{6}=84$.
Exercises
- 1 Use row $7$ to write row $8$.
- 2 Without writing it out, find the sum of the entries of row $9$.
- 3 Write row $9$ and name its largest entry. How many times does it appear?
- 4 Write $11=10+1$ and use row $4$ to evaluate $11^4$.
- 5 Row $5$ is $1,5,10,10,5,1$. Explain why $11^5$ is not $15101051$, and find its true value from the row.
- 6 Check from the rows that ${}^7C_3+{}^7C_4={}^8C_4$.
- 7 In which row does the number $15$ first appear? Which entry of that row is it, written as ${}^nC_r$?
- 1 $(a+b)^6$
- 2 $(x-2)^5$
- 3 $(2x+1)^4$
- 4 $(1-3x)^5$
- 5 $\left(\dfrac x3+\dfrac3x\right)^4$, $\ x\ne0$
- 6 $\left(x^2+\dfrac2x\right)^4$, $\ x\ne0$
- 7 $\left(2a-\dfrac b2\right)^5$
- 8 $\left(x+\dfrac1x\right)^6$, $\ x\ne0$
- 1 Show that $(a+b)^4-(a-b)^4=8a^3b+8ab^3$. Hence evaluate $(\sqrt3+1)^4-(\sqrt3-1)^4$.
- 2 Show that $(x+1)^6+(x-1)^6$ $=2\left(x^6+15x^4+15x^2+1\right)$. Hence evaluate $(\sqrt3+1)^6+(\sqrt3-1)^6$.
- 3 Evaluate $(\sqrt5+1)^5-(\sqrt5-1)^5$.
- 4 How many terms does the expansion of $(x+2y)^{12}$ have? Write its first three terms.
- 5 Use the binomial theorem and $i^2=-1$ (Chapter 4) to show that $(1+i)^4=-4$.
- 6 Find ${}^{10}C_0+{}^{10}C_1+\cdots+{}^{10}C_{10}$, and then ${}^{10}C_1+{}^{10}C_3+{}^{10}C_5+{}^{10}C_7+{}^{10}C_9$.
- 7 Expand $(a+b)^3$ and use it to show that $a^3+b^3=(a+b)^3-3ab(a+b)$. Hence find $a^3+b^3$ when $a+b=5$ and $ab=6$.
- 8 Prove that $2^n\ge n+1$ for every positive integer $n$.
- 1 Five coins are tossed. The number of outcomes with exactly $k$ heads is ${}^5C_k$. Write these numbers for $k=0,1,\dots,5$, check that they add up to the total number of outcomes, and find how many outcomes have at least $3$ heads.
- 2 A cube has edge $10.5$ cm. Write $10.5=10+0.5$ and use $(a+b)^3$ to find its volume.
- 3 A square metal plate of side $20$ cm is heated and each side grows by $0.1$ cm. Find the new area by expanding $(20+0.1)^2$. Which piece of the area model accounts for almost all of the increase?
- 4 A test has $6$ true/false questions. In how many ways can a student get exactly $4$ answers right? At least $4$ right?
- 5 A computer stores data as strings of $0$s and $1$s. How many strings of length $8$ contain exactly three $1$s? How many strings of length $8$ are there altogether?
How this is examined: Board papers ask you to expand a power such as $\left(x^2-\tfrac3x\right)^4$ in full (2โ3 marks), to evaluate sums like $(\sqrt2+1)^6+(\sqrt2-1)^6$, and to use ${}^nC_0+\cdots+{}^nC_n=2^n$. JEE goes straight for one term of the expansion, using the general term from the ๐ box.
E1. Board ยท 1 mark The number of terms in the expansion of $(2x-y)^9$ is (A) $9$ (B) $10$ (C) $18$ (D) $8$
E2. Board ยท 2 marks Expand $\left(x^2-\dfrac3x\right)^4$, $x\ne0$.
E3. Board ยท 3 marks Find the value of $(\sqrt2+1)^6+(\sqrt2-1)^6$.
E4. Board ยท Case study A door lock in a school laboratory is opened by a code of $6$ switches, each set to $0$ (down) or $1$ (up). (i) How many codes are possible? (1 mark) (ii) How many codes have exactly two switches up? (1 mark) (iii) Using the binomial theorem, show that exactly half of the codes have an even number of switches up. (2 marks)
E5. JEE Main The coefficient of $x^5$ in the expansion of $(1+2x)^8$ is (A) $56$ (B) $448$ (C) $1792$ (D) $3584$
E6. JEE Advanced Numerical answer. How many terms of the expansion of $\left(\sqrt2+\sqrt[3]{3}\right)^{12}$ are rational numbers?
Research question. If you toss four coins again and again, how often do you get $0$, $1$, $2$, $3$ or $4$ heads โ and does Pascal's triangle predict it?
Why it matters. The same counting lies behind the "bean machine" (Galton board) in science museums, where balls bounce left or right off rows of pins, and behind quality control in factories, where an inspector counts faulty items in a batch. Each outcome of four tosses is a choice of H or T for every coin, so โ exactly as in the proof of Theorem 7.1 โ the number of outcomes with $r$ heads is ${}^4C_r$, out of $2^4=16$ equally likely outcomes.
What to do.
- Predict: out of $64$ rounds of four tosses, how many rounds should give $0,1,2,3,4$ heads? (Multiply row $4$ by $\tfrac{64}{16}$.)
- Toss four coins $64$ times (or share the work: four friends do $16$ rounds each) and tally the number of heads in each round.
- Draw a bar chart of your counts next to the predicted counts.
- Repeat with a phone's random-number generator or a spreadsheet formula such as
=RANDBETWEEN(0,1)for $640$ rounds, and compare the shape again.
Tools: Four coins, a spreadsheet or random-number app
Hand in: Your tally table, the two bar charts, and three sentences on how close the experiment came to the prediction and why the large simulation is closer.
Going further. Do the same with six coins and row $6$. What fraction of rounds should have exactly three heads?
๐งช Self-check quiz
Q1. How many terms are there in the expansion of $(a+b)^{10}$?
Q2. $(1-2x)^4$ equals
Q3. The sum of the entries in row $7$ of Pascal's triangle is
Q4. In the counting proof of the binomial theorem, the coefficient of $a^{n-r}b^r$ is ${}^nC_r$ because
Q5. $(\sqrt2+1)^4+(\sqrt2-1)^4$ equals