Binomial Theorem

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XI, Chapter 7.

Can you find 98 to the power 5 without a calculator?

Multiplying 98 by itself five times is slow and error-prone. But 98 is just $100 - 2$, and powers of 100 are trivial. The Binomial Theorem is the machine that turns $(100-2)^5$ into five easy terms. It also shows in a line that $(1.01)^{1000000}$ is larger than 10000, which no calculator would tell you at a glance. This chapter is about expanding $(a+b)^n$ for a positive whole number $n$, and about reading information directly out of the expansion.

Spot the pattern first

Expand the small powers by hand and look.

Three things never change. There are one more term than the index. The power of $a$ drops by 1 and the power of $b$ rises by 1 as you move right. And in every term the two powers add up to $n$.

The coefficients form a triangle: 1 at the top, and every other entry is the sum of the two entries above it. This is Pascal's triangle, known in India as Meru Prastara.

The triangle works for small $n$, but you would need every earlier row to reach row 20. We need a direct formula for the entries.

Beyond the textbook. For the exam: the triangle is Pascal's triangle, also called Meru Prastara. In reality: this array of numbers was described in Indian texts on Sanskrit poetic metre more than two thousand years ago, long before Pascal studied it in the seventeenth century. It has been discovered independently in several countries, which is why different cultures attach different names to the same numbers.

Why the coefficients are combinations

Write $(a+b)^n$ as $n$ identical brackets multiplied together: $(a+b)(a+b)\cdots(a+b)$. To build a term, pick either $a$ or $b$ from each bracket. For the term $a^{n-k}b^{k}$ you must pick $b$ from exactly $k$ of the brackets. The number of ways to choose which $k$ brackets give the $b$ is $^{n}C_{k}$. That is the coefficient.

This is the Binomial Theorem for a positive integer $n$:

$$(a+b)^n = {}^{n}C_{0}\,a^{n} + {}^{n}C_{1}\,a^{n-1}b + {}^{n}C_{2}\,a^{n-2}b^{2} + \cdots + {}^{n}C_{n}\,b^{n} = \sum_{k=0}^{n} {}^{n}C_{k}\,a^{n-k}b^{k}$$

The formal proof uses mathematical induction. The step from $n=k$ to $n=k+1$ depends on Pascal's identity $^{k}C_{r} + {}^{k}C_{r-1} = {}^{k+1}C_{r}$, which is exactly why each triangle entry is the sum of the two above it. The numbers $^{n}C_{r}$ are called binomial coefficients.

Worked example: expanding $(x - 2y)^5$. Treat it as $(x + (-2y))^5$.

Common wrong step: writing every term with a plus sign. The signs alternate because $b = -2y$ is negative: odd powers of $b$ give minus, even powers give plus.

The three special cases and what they tell you

Putting $b=-y$, or $a=1$ and $b=\pm x$, gives expansions you will use constantly.

Now put $x=1$ in each of the last two:

$$ {}^{n}C_{0} + {}^{n}C_{1} + \cdots + {}^{n}C_{n} = 2^{n}, \qquad {}^{n}C_{0} - {}^{n}C_{1} + {}^{n}C_{2} - \cdots + (-1)^{n}\,{}^{n}C_{n} = 0$$

Adding and subtracting these two facts shows that the coefficients in even positions add to $2^{n-1}$, and so do those in odd positions. This trick of substituting a convenient value for $x$ is the fastest way to find the sum of coefficients of any expansion. For instance, the sum of the coefficients of $(3x-2)^4$ is found by putting $x=1$: $(3-2)^4 = 1$.

Pause and predict: how many terms survive in $(x+y)^6 + (x-y)^6$?

Check your answer: 4. The odd powers of $y$ appear with opposite signs in the two expansions and cancel, while the even powers double. You get $2\left(x^6 + 15x^4y^2 + 15x^2y^4 + y^6\right)$.

The general term: the one formula to master

In $(a+b)^n$ the term with $b^r$ is the $(r+1)$-th term:

$$T_{r+1} = {}^{n}C_{r}\,a^{\,n-r}\,b^{\,r}$$

The position is $r+1$ because counting starts at $r=0$. Almost every "find the term" question reduces to writing $T_{r+1}$, simplifying the power of $x$, and solving for $r$.

Worked example: a coefficient. Find the coefficient of $x^3$ in $(2+3x)^6$.

Common wrong step: reporting $^{6}C_{3}=20$. The coefficient of $x^3$ in $(2+3x)^6$ includes the numbers $2^3$ and $3^3$, not just the binomial coefficient. It equals $^{n}C_{r}$ only when $a=1$ and $b=x$.

Worked example: the term independent of $x$. Find the term independent of $x$ in $\left(x^{2} + \dfrac{1}{x}\right)^{9}$.

Common wrong step: setting $18-2r=0$. The factor $\left(\dfrac1x\right)^r$ also contributes $x^{-r}$, so both parts count.

Middle terms and the greatest coefficient

The expansion has $n+1$ terms, so the middle depends on whether $n+1$ is odd or even.

Worked example: two middle terms. Find the middle terms of $(2x - y)^5$.

Common wrong step: giving a single middle term for odd $n$. Whenever $n$ is odd, both $T_{(n+1)/2}$ and $T_{(n+3)/2}$ are middle terms.

The greatest binomial coefficient in row $n$ is the one in the middle, $^{n}C_{n/2}$ for even $n$. In row 10 it is $^{10}C_{5} = 252$. For odd $n$ the two middle coefficients are equal and both are greatest.

Using the theorem to calculate and to prove

Large powers. Break the number into a round number and a small adjustment.

Worked example: $98^5$.

Common wrong step: adding all terms. That is the expansion of $(100+2)^5 = 102^5$, not $98^5$.

Comparing sizes. To show $(1.01)^{1000000} > 10000$, write $(1+0.01)^{1000000}$ and keep the first two terms: $1 + 1000000 \times 0.01 = 10001$. Every other term is positive, so the whole is greater than 10001, hence greater than 10000. Dropping terms is legitimate only because they are positive.

Pause and predict: without a calculator, is $(1.01)^{100}$ greater than 2?

Check your answer: yes. The first two terms give $1 + 100 \times 0.01 = 2$, and all the remaining terms are positive, so the value is more than 2. (It is about 2.70.)

Divisibility. The trick is to write the number as $(1 + \text{something divisible by the divisor})^n$.

Worked example: remainder on division by 25. Show that $6^n - 5n$ leaves remainder 1 when divided by 25.

Common wrong step: splitting 6 as $2+4$ or $3+3$. The split must produce a factor of the divisor in every term after the first two, and $5^2=25$ does exactly that.

Beyond the textbook. For the exam: $(1+x)^n \approx 1 + nx$ is used only to compare numbers. In reality: the same first-order approximation is used in finance, physics and engineering when $x$ is small. For example, $(1.02)^5 \approx 1 + 5 \times 0.02 = 1.10$, while the exact value is $1.1041$. The error comes from the dropped terms, which are all positive, so for positive $x$ the shortcut always underestimates.

Beyond the textbook. For the exam: $\sum {}^{n}C_{k}=2^{n}$ is a result to remember. In reality: it is the reason probabilities add up to 1 when you toss a fair coin $n$ times. The chance of exactly $k$ heads is $^{n}C_{k}/2^{n}$. For 4 tosses the numbers are $1, 4, 6, 4, 1$ out of 16, a row of Pascal's triangle divided by $2^4$. This is the idea behind the binomial distribution you will meet in Class XII.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Permutations and Combinations gives $^{n}C_{r}$, the symmetry rule and Pascal's identity that drive the theorem.
• Next: Sequences and Series studies sums of terms, and the expansion is itself a finite series.
• Class XI link: Probability uses the same counting ideas for coin tosses and favourable outcomes.
• Class XII link: Probability uses the binomial distribution, built directly from this expansion.