Binomial Theorem
Expanding \((x+y)^n\) without multiplying it out — where counting meets algebra
- The binomial theorem for a positive integral index and why its coefficients are \({}^{n}C_r\).
- The general term \(T_{r+1}\) — the key that unlocks almost every problem.
- How to find the middle term, the term independent of \(x\), and the numerically greatest term.
- The properties of binomial coefficients and the famous coefficient sums.
- The binomial theorem for any index (negative or fractional) and the multinomial extension.
- Using expansions for approximation and divisibility / remainder arguments.
Why a Theorem for \((x+y)^n\)?
You can square a binomial by hand, and cube it with patience — but expanding \((x+y)^{20}\) by repeated multiplication is hopeless. Yet there is perfect order hiding in the mess. Multiplying out \((x+y)^n\) means choosing, from each of the \(n\) factors, either an \(x\) or a \(y\). The number of ways to land exactly \(r\) copies of \(y\) is precisely \({}^{n}C_r\) — so combinatorics from the previous chapter is the engine of this one.
The Binomial Theorem (Positive Integral Index)
Written out: \((x+y)^n={}^{n}C_0x^n+{}^{n}C_1x^{n-1}y+\dots+{}^{n}C_n y^n\), valid for every positive integer \(n\) and all real (or complex) \(x,y\).
Three features deserve attention from the start. There are exactly \(n+1\) terms. In each term the exponents of \(x\) and \(y\) add to \(n\). And the coefficients are symmetric, because \({}^{n}C_r={}^{n}C_{n-r}\).
| Form | Expansion |
|---|---|
| \((1+x)^n\) | \({}^{n}C_0+{}^{n}C_1x+{}^{n}C_2x^2+\dots+{}^{n}C_n x^n\) |
| \((x+y)^n\) | \(\sum_{r=0}^{n}{}^{n}C_r\,x^{n-r}y^{r}\) |
| \((x-y)^n\) | \(\sum_{r=0}^{n}(-1)^{r}\,{}^{n}C_r\,x^{n-r}y^{r}\) |
For \((x-y)^n\) the terms alternate in sign because of the factor \((-1)^r\). A dropped minus sign is the most common slip in the whole chapter — write the \((-1)^r\) explicitly before substituting numbers.
The General Term — Your Master Key
Almost no JEE problem asks you to write the whole expansion. Instead it asks for one term — the fifth term, the term in \(x^{7}\), the constant term. For all of these you only need the general term.
Note the index: \(T_{r+1}\) uses \(r\), so the 5th term needs \(r=4\). To find a specific power, set the exponent equal to the target and solve for \(r\).
The Middle Term(s)
With \(n+1\) terms, whether there is one middle term or two depends on the parity of \(n\). The middle term carries the largest binomial coefficient, which is why it appears so often in problems.
| Case | Middle term(s) |
|---|---|
| \(n\) even | one term: \(T_{\frac{n}{2}+1}\) |
| \(n\) odd | two terms: \(T_{\frac{n+1}{2}}\) and \(T_{\frac{n+3}{2}}\) |
Picking Out a Particular Term
Three classic tasks, all solved by the general term.
Write \(T_{r+1}\), collect the total power of \(x\), set it equal to \(k\), solve for \(r\), then read the coefficient.
Same idea with target power \(0\): set the exponent of \(x\) to zero. If no integer \(r\) works, no constant term exists.
Use the ratio \(\left|\dfrac{T_{r+1}}{T_r}\right|\ge 1\) to find where terms stop growing.
This gives \(r\le\dfrac{(n+1)|x|}{1+|x|}\). The greatest term is at the largest integer \(r\) satisfying it; if that bound is itself an integer, two consecutive terms are equal and both are greatest.
Binomial Coefficients & Pascal's Triangle
The numbers \({}^{n}C_0,{}^{n}C_1,\dots,{}^{n}C_n\) are the binomial coefficients, usually abbreviated \(C_0,C_1,\dots,C_n\) for a fixed \(n\). Stacked by rows they form Pascal's triangle, where each entry is the sum of the two above it — exactly Pascal's rule \({}^{n}C_r+{}^{n}C_{r-1}={}^{n+1}C_r\).
| Identity | Meaning / use |
|---|---|
| \({}^{n}C_r={}^{n}C_{n-r}\) | symmetry — coefficients read the same backwards |
| \({}^{n}C_r+{}^{n}C_{r-1}={}^{n+1}C_r\) | Pascal's rule — builds the triangle |
| \(\dfrac{{}^{n}C_r}{{}^{n}C_{r-1}}=\dfrac{n-r+1}{r}\) | ratio of consecutive terms — drives greatest-term and series work |
| \(r\,{}^{n}C_r=n\,{}^{n-1}C_{r-1}\) | absorbs a factor of \(r\) — used to sum \(\sum r\,C_r\) |
Sums of Binomial Coefficients
Substituting clever values of \(x\) into \((1+x)^n\) turns the theorem into a coefficient-summing machine. These results are worth memorising.
The sum of squares \(C_0^2+C_1^2+\dots+C_n^2={}^{2n}C_n\) comes from comparing the coefficient of \(x^n\) on both sides of \((1+x)^n(1+x)^n=(1+x)^{2n}\).
Binomial Theorem for Any Index & the Multinomial
When the index is not a positive integer, the expansion never terminates — it becomes an infinite series, valid only when the variable is small.
Here \(n\) may be negative or fractional and the series is infinite. It converges only for \(|x|<1\). Note \({}^{n}C_r\) in the integer case is replaced by the falling product \(\dfrac{n(n-1)\cdots(n-r+1)}{r!}\).
| Series | Expansion |
|---|---|
| \((1-x)^{-1}\) | \(1+x+x^2+x^3+\cdots\) |
| \((1+x)^{-1}\) | \(1-x+x^2-x^3+\cdots\) |
| \((1-x)^{-2}\) | \(1+2x+3x^2+4x^3+\cdots\) |
For three or more terms, \((x_1+x_2+\dots+x_k)^n=\displaystyle\sum\dfrac{n!}{r_1!\,r_2!\cdots r_k!}\,x_1^{r_1}x_2^{r_2}\cdots x_k^{r_k}\), the sum taken over all non-negative integers with \(r_1+r_2+\dots+r_k=n\). The number of distinct terms is \({}^{\,n+k-1}C_{\,k-1}\) — stars and bars from Chapter 5.
Applications: Approximation & Divisibility
The binomial theorem is not only an algebraic identity — it is a practical tool.
For small \(x\), \((1+x)^n\approx 1+nx\). So \((1.002)^{10}\approx 1+10(0.002)=1.02\) — accurate to three decimals.
To study \(a^n\bmod m\), write \(a\) as \((m\pm k)\) and expand: every term but the last carries a factor of \(m\).
Strategies & Standard Results
A short checklist that resolves most binomial problems.
Write the general term, simplify the power of the variable, then impose the condition the problem asks for.
For coefficient sums, plug \(x=1,-1,\) or \(i\) into \((1+x)^n\); differentiate or integrate first if the sum is weighted.
To prove a coefficient identity, expand two sides of a product (e.g. \((1+x)^a(1+x)^b\)) and match the same power of \(x\).
Putting It to Work
Problem. Find the 6th term in the expansion of \(\left(2x-\dfrac{1}{x}\right)^{9}\).
Solution. The 6th term means \(r=5\):
Problem. Find the constant term in \(\left(x^{2}+\dfrac{1}{x}\right)^{9}\).
Solution. General term \(T_{r+1}={}^{9}C_r(x^2)^{9-r}\left(\tfrac1x\right)^r={}^{9}C_r\,x^{18-3r}\). Set the exponent to zero: \(18-3r=0\Rightarrow r=6\).
Problem. Find the middle term of \(\left(\dfrac{x}{2}+2y\right)^{8}\).
Solution. Here \(n=8\) is even, so there is a single middle term, the \(\left(\tfrac{8}{2}+1\right)=5\)th, i.e. \(r=4\):
Problem. Find the coefficient of \(x^{5}\) in \(\left(x+\dfrac{3}{x^{2}}\right)^{14}\).
Solution. \(T_{r+1}={}^{14}C_r\,x^{14-r}\left(\tfrac{3}{x^2}\right)^{r}={}^{14}C_r\,3^{r}\,x^{14-3r}\). Set \(14-3r=5\Rightarrow r=3\).
Problem. Evaluate \(C_0+2C_1+3C_2+\dots+(n+1)C_n\).
Solution. Split as \(\sum_{r=0}^{n}(r+1)C_r=\sum rC_r+\sum C_r\). The first sum is \(n\,2^{n-1}\) and the second is \(2^{n}\).
Problem. Show that \(3^{2n+2}-8n-9\) is divisible by \(64\).
Solution. Write \(3^{2n+2}=9^{\,n+1}=(1+8)^{\,n+1}\) and expand:
Subtracting \(8n+9\) leaves only terms containing \(8^{2}=64\), so the expression is divisible by 64.
Chapter Summary
\((x+y)^n=\sum {}^{n}C_r x^{n-r}y^r\), with \(n+1\) terms and exponents summing to \(n\).
\(T_{r+1}={}^{n}C_r x^{n-r}y^r\) — set \(r\) to extract any term; the master key.
One term \(T_{n/2+1}\) if \(n\) even; two if \(n\) odd. Largest coefficient sits here.
\(\sum C_r=2^n\), even = odd = \(2^{n-1}\), \(\sum C_r^2={}^{2n}C_n\).
\((1+x)^n=1+nx+\tfrac{n(n-1)}{2!}x^2+\cdots\) — infinite, valid for \(|x|<1\).
Approximate with \(1+nx\); settle divisibility by expanding \((m\pm k)^n\).
Problems
For each, write the general term first. Difficulty rises down the list.
- Find the 7th term in the expansion of \(\left(3x^{2}-\dfrac{1}{3x}\right)^{9}\).
- Find the term independent of \(x\) in \(\left(2x^{2}-\dfrac{1}{x}\right)^{12}\).
- Find the middle term(s) of \(\left(x-\dfrac{1}{x}\right)^{11}\).
- Find the coefficient of \(x^{9}\) in \((1+3x+3x^{2}+x^{3})^{6}\). (Hint: it is \((1+x)^{18}\).)
- If the coefficients of the 5th, 6th and 7th terms of \((1+x)^n\) are in arithmetic progression, find \(n\).
- Find the numerically greatest term in \((3+2x)^{15}\) when \(x=\tfrac{1}{5}\).
- Prove that \(C_0^2+C_1^2+\dots+C_n^2={}^{2n}C_n\) by comparing coefficients in \((1+x)^{n}(1+x)^{n}\).
- Evaluate \(\dfrac{C_0}{1}+\dfrac{C_1}{2}+\dfrac{C_2}{3}+\dots+\dfrac{C_n}{n+1}\).
- Find the remainder when \(7^{103}\) is divided by \(25\).
- Using the binomial series, approximate \((0.998)^{8}\) to four decimal places.
- How many distinct terms appear in the expansion of \((x+y+z)^{10}\), and what is the coefficient of \(x^{4}y^{3}z^{3}\)?
- Show that \(\big(\sqrt{2}+1\big)^{6}+\big(\sqrt{2}-1\big)^{6}\) is an integer, and find it.