JEE Advanced (IIT Entrance) · Mathematics (JEE & NDA)
Binomial Theorem and Its Simple Applications
Binomial expansion for positive integral index, general and middle terms, and simple applications.
Six concepts for the positive-integral case JEE Main actually asks: the expansion, the general term (and the term without x), middle terms, coefficient sums, the greatest term, and the linear approximation (1+x)^n\approx 1+nx.
- JEE Advanced (IIT Entrance)
- Easy level
- 6 concepts
- 5 practice questions
1Binomial expansion for positive integral n
For a positive integer n, (a+b)^n=\sum_{r=0}^{n}{}^nC_r\,a^{n-r}b^r. The sum runs from r=0 to r=n, so there are n+1 terms, not n. The usual slip is counting the index as the term count. The coefficients {}^nC_r are symmetric about the middle: {}^nC_r={}^nC_{n-r}, which is why the Pascal row reads the same forwards and backwards.
Write the sum \sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^{r} term by term — the figure would only re-typeset that identity.
Reading the expansion
- Write the sum(a+b)^n expands as \sum_{r=0}^{n}{}^nC_r\,a^{n-r}b^r — one term for each r.
- Count termsr runs through n+1 values, so the expanded polynomial has n+1 terms.
- Use symmetry{}^nC_r={}^nC_{n-r} pairs the ends of the row; the greatest coefficient sits in the middle.
| Piece | Role |
|---|---|
| n | Positive integer index; fixed for the expansion |
| r | Term index in the sum; runs 0,1,\ldots,n |
| {}^nC_r | Binomial coefficient of the (r+1)-th term |
| a^{n-r}b^r | Powers: a falls as b rises, total degree n |
The expansion of (a+b)^{10} has how many terms?
- 10
- 11
- 9
r runs from 0 to 10, which is eleven values. Answering 10 counts the index instead of the terms; answering 9 drops both endpoints.
2General term, and the term independent of x
The (r+1)-th term of (a+b)^n is T_{r+1}={}^nC_r\,a^{n-r}b^r. The term number is one ahead of the exponent index, because the exponent index starts at zero: T_1 is the a^n term. To locate a specific power of x, or the term independent of x, write T_{r+1}, collect the total exponent of x, set that exponent to the required value and solve for r.
Figure. Write T_{r+1}={}^9C_r\,x^{18-3r} for (x^2+1/x)^9. The independent term needs exponent 18-3r=0, so r=6, and {}^9C_6=84.
Finding a required term
- Write T_{r+1}Substitute the two parts of the binomial into T_{r+1}={}^nC_r\,a^{n-r}b^r.
- Collect the power of xSimplify every factor that carries x into a single power x^{\ldots}.
- Solve for rSet the exponent to the required value (zero for the independent term) and evaluate {}^nC_r.
Term independent of x
Find the term independent of x in the expansion of \left(x^2+\dfrac{1}{x}\right)^9.
- T_{r+1}={}^9C_r\,(x^2)^{9-r}\left(\dfrac{1}{x}\right)^r{}^9C_r\,x^{18-3r}
- Set power of x to zero: 18-3r=0r=6
- Term ={}^9C_6={}^9C_3=\dfrac{9\cdot8\cdot7}{3\cdot2\cdot1}84
Pro tip. Set the combined exponent of x to zero; the value of r you get plugs straight into {}^nC_r. Do not forget that T_{r+1} means the power on b is r, not r+1.
In T_{r+1} for (a+b)^n, the power of b is
- r+1
- r
- n-r
T_{r+1}={}^nC_r\,a^{n-r}b^r, so b carries exponent r. Choosing r+1 confuses the term number with the exponent; n-r is the power of a.
3Middle term(s)
The middle of the expansion depends on whether n is even or odd. If n is even there is one middle term, the \left(\dfrac{n}{2}+1\right)-th. If n is odd there are two middle terms, the \left(\dfrac{n+1}{2}\right)-th and the \left(\dfrac{n+3}{2}\right)-th. The trap is treating every expansion as if it had a single middle term.
Middle means an index in a finite list of n+1 terms — odd n gives one middle term, even n gives two.
| n | Middle term(s) |
|---|---|
| Even | One: T_{\frac{n}{2}+1} |
| Odd | Two: T_{\frac{n+1}{2}} and T_{\frac{n+3}{2}} |
Middle term of an expansion
Find the middle term in the expansion of (2x-3)^6.
- n=6 is even → single middle term at \dfrac{6}{2}+1T_4
- T_4={}^6C_3(2x)^3(-3)^320\times 8x^3\times(-27)
- 20\times 8\times(-27)\,x^3-4320x^3
Pro tip. For even n there is exactly one middle term at position \tfrac{n}{2}+1; keep track of the sign when a factor is negative — here (-3)^3 supplies the minus.
In the expansion of (a+b)^7, the number of middle terms is
- One
- Two
- None — middle terms are defined only for even n
n=7 is odd, so the middle is shared by T_4 and T_5. The third option invents a restriction the definition does not have; the first option is the even-n habit applied to an odd index.
4Sums of binomial coefficients
Substituting specific values of a and b turns the expansion into identities about the coefficients alone. Put a=b=1 and the left side is 2^n, so the sum of all coefficients is 2^n. Put a=1, b=-1 and the alternating sum is 0; combining that with the full sum isolates the even-position and odd-position sums, each equal to 2^{n-1}. The greatest coefficient in the row is the middle one.
Figure. Put a=b=1 to get every coefficient summing to 2^n; for (x+y)^8 that is 256. Put a=1, b=-1 to get alternating sum 0. Adding those identities splits the row into even-position and odd-position sums, each 2^{n-1}.
Isolating even and odd sums
- Full sumSet a=b=1: (1+1)^n=2^n=\sum{}^nC_r.
- Alternating sumSet a=1, b=-1: (1-1)^n=0=\sum(-1)^r{}^nC_r.
- Add or subtractAdding the two identities doubles the even-position sum; subtracting doubles the odd-position sum. Each is 2^{n-1}.
Even- and odd-position sums
From (a+b)^n, show that the sum of the even-position coefficients equals the sum of the odd-position coefficients, and find each sum.
- (1+1)^n\sum{}^nC_r=2^n
- (1-1)^n\sum(-1)^r{}^nC_r=0
- Add the two identitieseven-position sum = odd-position sum =2^{n-1}
Pro tip. Substitute x=1 for the sum of all coefficients and x=-1 for the alternating sum; combining them isolates even- or odd-position sums. "Position" here follows the r-index in {}^nC_r, starting at r=0.
The sum of the coefficients in the expansion of (x+y)^8 is
- 8
- 256
- 128
Put x=y=1: (1+1)^8=2^8=256. Answering 8 reports the index; 128=2^7 is each of the even- or odd-position sums, not the full sum.
5Greatest term in the expansion
"Greatest coefficient" and "greatest term" are different questions. The greatest coefficient is the middle {}^nC_r, settled by symmetry. The greatest term also depends on the ratio b/a: compare consecutive terms by \dfrac{T_{r+1}}{T_r}=\dfrac{n-r+1}{r}\cdot\dfrac{b}{a} and find where that ratio crosses 1. Terms grow while the ratio is greater than 1 and shrink once it falls below 1.
Figure. Greatest term is not the same as greatest coefficient. Form T_{r+1}/T_r=((n-r+1)/r)\cdot(b/a) and read the peak where that ratio crosses 1: terms grow while the ratio is above 1 and shrink once it falls below.
Where the terms peak
- Form the ratioWrite \dfrac{T_{r+1}}{T_r}=\dfrac{n-r+1}{r}\cdot\dfrac{b}{a} from the general term.
- Solve \dfrac{T_{r+1}}{T_r}\ge 1The inequality marks every r for which the next term is at least as large as the current one.
- Read the peakThe greatest term (or two equal greatest terms) sits where the ratio crosses 1.
To find the greatest term (not merely the greatest coefficient) in (a+b)^n, you must use
- Only the middle index \tfrac{n}{2}+1, because the largest {}^nC_r always wins
- The ratio \dfrac{T_{r+1}}{T_r} and where it crosses 1
- Only the sum of coefficients 2^n
The middle coefficient is largest, but a large |b/a| can push the greatest term away from the middle. The ratio test is what tracks that. 2^n is the sum of coefficients and says nothing about which single term is biggest.
6Linear approximation (1+x)^n\approx 1+nx
When |x| is much smaller than 1, the binomial expansion of (1+x)^n is dominated by its first two terms: (1+x)^n\approx 1+nx. Higher powers x^2, x^3, \ldots are negligible beside x only while that increment stays tiny. The same expansion, written with a convenient multiple plus remainder — (\text{multiple}+r)^n — is the standard route to last digits and remainders on exam papers.
Figure. Near x=0, (1+x)^3 and its linearisation 1+3x share value and slope; the gap grows as |x| leaves the tiny-increment regime where higher powers stay negligible.
| Kept | Dropped |
|---|---|
| 1 | — |
| nx | \dfrac{n(n-1)}{2}x^2 and higher |
The approximation (1+x)^n\approx 1+nx is used when
- |x| is much smaller than 1
- n is even
- x=1, so the right side becomes 1+n
The higher powers of x are negligible only for |x|\ll 1. Even n is about middle terms, not this truncation. At x=1 the exact value is 2^n, while 1+n is nowhere near it for large n — so x=1 is exactly when the approximation must not be used.
Notes
- Binomial theorem: For a positive integer n, (a+b)^n=\sum_{r=0}^{n}{}^nC_r\,a^{n-r}b^r. There are n+1 terms and the coefficients {}^nC_r are symmetric about the middle.
- General term: The (r+1)-th term is T_{r+1}={}^nC_r\,a^{n-r}b^r. Setting the power of the variable to a required value locates a specific term or the term independent of x.
- Middle term: If n is even there is one middle term, the \left(\dfrac{n}{2}+1\right)-th; if n is odd there are two middle terms, the \left(\dfrac{n+1}{2}\right)-th and \left(\dfrac{n+3}{2}\right)-th.
- Properties of coefficients: The sum of all coefficients is 2^n (put a=b=1); the sum of even-position coefficients equals the sum of odd-position coefficients, each 2^{n-1}. The greatest coefficient is the middle one.
- Applications: Binomial expansion approximates (1+x)^n\approx1+nx for small x, and helps find remainders and last digits by writing a number as (\text{multiple}+r)^n.
- Differentiation and integration of (1+x)^n generate identities for \sum k\binom{n}{k} and \sum \binom{n}{k}/(k+1) - a favourite Advanced technique.
- The greatest term / greatest coefficient is found by examining the ratio T_{r+1}/T_r and locating where it crosses 1.
- Multinomial theorem generalises the expansion; the coefficient of a specific monomial is a multinomial coefficient.
- For (1+x)^n with n negative or fractional, the series is infinite and valid only for |x|<1; keep the general term \binom{n}{r} with the generalized definition.
Formulas
- Expansion: (a+b)^n=\sum_{r=0}^{n}{}^nC_r\,a^{n-r}b^r
- General term: T_{r+1}={}^nC_r\,a^{n-r}b^r
- Sum of coefficients: put x=1 to get 2^n
- Even/odd coefficient sums: each =2^{n-1}
- Middle term (n even): T_{\frac{n}{2}+1}
- Approximation: (1+x)^n\approx1+nx for |x|\ll1
- \sum_{k=0}^{n} k\binom{n}{k} = n\,2^{n-1}
- \sum_{k=0}^{n} \dfrac{1}{k+1}\binom{n}{k} = \dfrac{2^{n+1}-1}{n+1}
- General term: T_{r+1}=\binom{n}{r}a^{n-r}b^{r}
Exam traps & shortcuts
- To find the term independent of x, write the general term, collect the total power of x, set it to zero and solve for r.
- Substitute x=1 to get the sum of all coefficients and x=-1 to get the alternating sum; combining them isolates even- or odd-position sums.
- For 'greatest term' problems, use the ratio \dfrac{T_{r+1}}{T_r}=\dfrac{n-r+1}{r}\cdot\dfrac{b}{a} and find where it crosses 1.
- To evaluate \sum k^2\binom{n}{k}, apply x\dfrac{d}{dx} twice to (1+x)^n then set x=1.
- For products of two binomial expansions, compare coefficients using Vandermonde's identity \sum_k\binom{m}{k}\binom{n}{p-k}=\binom{m+n}{p}.
Reference tables
Positive integral index only. Every identity below is the expansion with a particular substitution or a particular term picked out.
| Name | Statement |
|---|---|
| Expansion | (a+b)^n=\sum_{r=0}^{n}{}^nC_r\,a^{n-r}b^r |
| General term | T_{r+1}={}^nC_r\,a^{n-r}b^r |
| Sum of coefficients | Put a=b=1: sum =2^n |
| Even / odd sums | Each =2^{n-1} |
| Middle term (n even) | T_{\frac{n}{2}+1} |
| Approximation | (1+x)^n\approx 1+nx for |x|\ll 1 |
Recap
Read only this the night before.
- Expansion
- (a+b)^n=\sum{}^nC_r\,a^{n-r}b^r has n+1 terms. Coefficients are symmetric: {}^nC_r={}^nC_{n-r}.
- General term
- T_{r+1}={}^nC_r\,a^{n-r}b^r. For the term independent of x, set the total power of x to zero and solve for r.
- Middle
- Even n: one middle term at \tfrac{n}{2}+1. Odd n: two, at \tfrac{n+1}{2} and \tfrac{n+3}{2}.
- Coefficient sums
- All coefficients sum to 2^n. Even-position and odd-position sums are each 2^{n-1} — get them from x=1 and x=-1.
- Greatest term
- Use \dfrac{T_{r+1}}{T_r}=\dfrac{n-r+1}{r}\cdot\dfrac{b}{a} and find where it crosses 1. Middle coefficient ≠ greatest term when |b/a| is large.
- Approximation
- (1+x)^n\approx 1+nx only for |x|\ll 1. Remainders and last digits: write the base as (\text{multiple}+r)^n.
Practise Binomial Theorem and Its Simple Applications
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- 5 exam-style questions on this topic, with explanations
- A 5-question practice set that ends the chapter
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