**Binomial Theorem** – As the power increases the expansion becomes lengthy and tedious to calculate. A binomial expression that has been raised to a very large power can be easily calculated with the help of the Binomial Theorem. Learn about all the details about the binomial theorem like its definition, properties, applications, etc. and download the binomial theorem PDF lesson from below.

## Binomial Theorem Guide

- Introduction to the Binomial Theorem
- Properties of Binomial coefficients
- Terms in the Binomial Expansion
- Binomial Theorem for any Index
- Applications of Binomial Theorem
- Multinomial Theorem
- Problems on Binomial Theorem

**Download this lesson as PDF:-**Binomial Theorem PDF

## Introduction to the Binomial Theorem

The Binomial Theorem is the method of expanding an expression that has been raised to any finite power. A binomial Theorem is a powerful tool of expansion, which has application in Algebra, probability, etc.

**Binomial Expression: **A binomial expression is an algebraic expression that contains two dissimilar terms. Ex: a + b, a^{3} + b^{3}, etc.

**Binomial Theorem:** Let n ∈ N,x,y,∈ R then

(x + y)^{n} = ^{n}Σ_{r=0} nC_{r} x^{n – r }· y^{r} where,

**Illustration 1:** **Expand (x/3 + 2/y) ^{4}**

**Sol:**

**Illustration 2: (√2 + 1) ^{5} + (√2 − 1)^{5}**

**Sol:**

We have

(x + y)^{5} + (x – y)^{5} = 2[5C_{0} x^{5} + 5C_{2} x^{3} y^{2} + 5C_{4} xy^{4}]

= 2(x^{5 }+ 10 x^{3} y^{2 }+ 5xy^{4})

Now (√2 + 1)^{5 }+ (√2 − 1)^{5 }= 2[(√2)^{5 }+ 10(√2)^{3}(1)^{2 }+ 5(√2)(1)^{4}]

=58√2

## Binomial Expansion

Important points to remember

- The total number of terms in the expansion of (x+y)
^{n}are (n+1) - The sum of exponents of x and y is always n.
- nC
_{0}, nC_{1}, nC_{2}, … .., nC_{n}are called binomial coefficients and also represented by C_{0}, C_{1}, C_{2}, ….., C_{n} - The binomial coefficients which are equidistant from the beginning and from the ending are equal i.e. nC
_{0 }= nC_{n}, nC_{1 }= nC_{n-1 }, nC_{2 }= nC_{n-2},….. etc.

To find binomial coefficients we can also use **Pascal’s Triangle**.

**Some other useful expansions:**

- (x + y)
^{n }+ (x−y)^{n }= 2[C_{0}x^{n }+ C_{2}x^{n-1}y^{2 }+ C_{4}x^{n-4}y^{4 }+ …] - (x + y)
^{n }– (x−y)^{n }= 2[C_{1}x^{n-1 }y + C_{3}x^{n-3}y^{3 }+ C_{5}x^{n-5}y^{5 }+ …] - (1 + x)
^{n }=^{n}Σ_{r-0}nC_{r }. x^{r }= [C_{0 }+ C_{1}x + C_{2}x^{2 }+ … C_{n}x_{n}] - (1+x)
^{n }+ (1 − x)^{n }= 2[C_{0}+ C_{2}x^{2}+C_{4}x^{4 }+ …] - (1+x)
^{n }− (1−x)^{n }= 2[C_{1}x + C_{3}x^{3}+ C_{5}x^{5}+ …] - The number of terms in the expansion of (x + a)
^{n}+ (x−a)^{n}are (n+2)/2 if “n” is even or (n+1)/2 if “n” is odd. - The number of terms in the expansion of (x + a)
^{n}− (x−a)^{n }are (n/2) if “n” is even or (n+1)/2 if “n” is odd.

### Number of Terms and R-F Factor Relation

### Properties of Binomial Coefficients

Binomial coefficients refer to the integers which are coefficients in the binomial theorem. Some of the most important properties of binomial coefficients are:

- C
_{0 }+ C_{1 }+ C_{2 }+ … + C_{n }= 2^{n} - C
_{0 }+ C_{2 }+ C_{4 }+ … = C_{1 }+ C_{3 }+ C_{5 }+ … = 2^{n-1} - C
_{0 }– C_{1 }+ C_{2 }– C_{3 }+ … +(−1)^{n}. nC_{n }= 0 - nC
_{1 }+ 2.nC_{2 }+ 3.nC_{3 }+ … + n.nC_{n }= n.2^{n-1} - C
_{1 }− 2C_{2 }+ 3C_{3 }− 4C_{4 }+ … +(−1)^{n-1 }C_{n }= 0 for n > 1 - C
_{0}^{2 }+ C_{1}^{2 }+ C_{2}^{2 }+ …C_{n}^{2 }= [(2n)!/ (n!)^{2}]

**Illustration:** If (1 + x)^{15} = a_{0} + a_{1}x + . . . . . + a_{15} x^{15} then, find the value of

**Sol:**

= C_{1}/C_{0} + 2 C_{2}/C_{1}+ 3C_{3}/C_{2 }+ . . . . + 15 C_{15}/C_{14}

= 15 + 14 + 13 + . . . . . + 1 = [15(15+1)]/2 = 120

### Properties of Binomial Coefficients Video Lesson

## Terms in the Binomial Expansion

In binomial expansion, it is often asked to find the middle term or the general term. The different terms in the binomial expansion that are covered here include:

- General Term
- Middle Term
- Independent Term
- Determining a Particular Term
- Numerically greatest term
- Ratio of Consecutive Terms/Coefficients

### General Term in binomial expansion:

We have (x + y)^{n }= nC_{0} x^{n }+ nC_{1} x^{n-1 }. y + nC_{2} x^{n-2} . y^{2 }+ … + nC_{n} y^{n}

General Term = T_{r+1} = nC_{r} x^{n-r} . y^{r}

- General Term in (1 + x)
^{n}is nC_{r }x^{r} - In the binomial expansion of (x + y)
^{n}, the r^{th}term from end is (n – r + 2)^{th}.

**Illustration:** Find the number of terms in (1 + 2x +x^{2})^{50}

**Sol: **

(1 + 2x + x^{2})^{50 }= [(1 + x)^{2}]^{50 }= (1 + x)^{100}

The number of terms = (100 + 1) = 101

**Illustration: **Find the fourth term from the end in the expansion of (2x – 1/x^{2})^{10}

**Sol:**

Required term =T_{10 – 4 + 2} = T_{8} = 10C_{7 }(2x)^{3 }(−1/x^{2})^{7 }= −960x^{-11}

### Middle Term(S) in the expansion of (x+y)^{ n.}^{n}

- If n is even then (n/2 + 1) Term is the middle Term.
- If n is odd then [(n+1)/2]
^{th }and [(n+3)/2)^{th}terms are the middle terms.

**Illustration: **Find the middle term of (1 −3x + 3x^{2 }– x^{3})^{2n}

**Sol:**

(1 − 3x + 3x^{2 }– x^{3})^{2n} = [(1 − x)^{3}]^{2n }= (1 − x)^{6n}

Middle Term = [(6n/2) + 1] term = 6nC_{3n} (−x)^{3n}

Determining a Particular Term:

- In the expansion of(ax
^{p}+ b/x^{q})^{n }the coefficient of x^{m}is the coefficient of T_{r+1}where r = [(np−m)/(p+q)] - In the expansion of (x + a)
^{n}, T_{r+1}/T_{r}= (n – r + 1)/r . a/x

**General and Middle Terms of Binomial Expansion**

### Independent Term

The term Independent of in the expansion of [ax^{p }+ (b/x^{q})]^{n} is

T_{r+1} = ^{n}C_{r} a^{n-r }b^{r}, where r = (np/p+q) (integer)

**Illustration:** Find the independent term of x in (x+1/x)^{6}

**Sol:**

r = [6(1)/1+1] = 3

The independent term is 6C_{3 }= 20

**Illustration:** Find the independent term in the expansion of:

**Sol: **

(x^{1/3} + 1 – 1 – 1/√x)^{10} = (x^{1/3} – 1/√x)^{10}

r = [10(1/3)]/[1/3+1/2] = 4

**∴**** T _{5 }= ^{10}C_{4} = 210**

### Numerically greatest term in the expansion of (1+x)^{n}:

- If [(n+1)|x|]/[|x|+1] = P, is a positive integer then P
^{th}term and (P+1)^{th}terms are numerically the greatest terms in the expansion of (1+x)^{n} - If[(n+1)|x|]/[|x|+1] = P + F, where P is a positive integer and 0 < F < 1 then (P+1)
^{th}term is numerically the greatest term in the expansion of (1+x)^{n}.

**Illustration: **Find the numerically greatest term in (1-3x)^{10} when x = (1/2)

**Sol:**

[(n + 1)|α|] / [|α| + 1] = (11 × 3/2)/(3/2+1) = 33/5 = 6.6

Therefore, T_{7} is the numerically greatest term.

T_{6 + 1} = 10C_{6 }. (−3x)^{6 }= 10C_{6} . (3/2)^{6}

### Ratio of Consecutive Terms/Coefficients:

Coefficients of x^{r} and x^{r + 1} are nC_{r – 1} and nC_{r} respectively.

(nC_{r} / nC_{r – 1}) = (n – r + 1) / r

**Illustration:** If the coefficients of three consecutive terms in the expansion of (1+x)^{n} are in the ratio 1:7:42 then find the value of n.

**Sol:**

Let (r – 1)^{th}, (r)^{th} and (r + 1)^{th} be the three consecutive terms.

Then, the given ratio is 1:7:42

Now (nC_{r-2} / nC_{r – 1}) = (1/7)

(nC_{r-2} / nC_{r – 1}) = (1/7) ⇒ [(r – 1)/(n − r+2)] = (1/7) ⇒ n−8r+9=0 → (1)

And,

(nC_{r-1} / nC_{r}) = (7/42) ⇒ [(r)/(n – r +1)] =(1/6) ⇒ n−7r +1=0 → (2)

From (1) & (2), n = 55

## Applications of Binomial Theorem

Binomial theorem has a wide range of applications in Mathematics like finding the remainder, finding digits of a number, etc. The most common binomial theorem applications are:

### Finding Remainder using Binomial Theorem

**Illustration: **Find the remainder when 7^{103} is divided by 25

**Sol:**

(7^{103 }/ 25) = [7(49)^{51} / 25)] = [7(50 − 1)^{51 }/ 25]

= [7(25K − 1) / 25] = [(175K – 25 + 25−7) / 25]

= [(25(7K − 1) + 18) / 25]

∴ The remainder = 18.

**Illustration:** If the fractional part of the number (2^{403} / 15) is (K/15), then find K.

**Sol:**

(2^{403} / 15) = [2^{3 }(2^{4})^{100} / 15]

= 8/15 (15 + 1)^{100} = 8/ 15 (15λ + 1) = 8λ + 8/15

∵ 8λ is an integer, fractional part = 8/15

So, K = 8.

### Finding Digits of a Number

**Illustration: **Find the last two digits of the number (13)^{10}

**Sol:**

(13)^{10 }= (169)^{5 }= (170 − 1)^{5}

= 5C_{0} (170)^{5 }− 5C_{1 }(170)^{4 }+ 5C_{2 }(170)^{3 }− 5C_{3 }(170)^{2 }+ 5C_{4 }(170) − 5C_{5}

= 5C_{0 }(170)^{5 }− 5C_{1 }(170)^{4 }+ 5C_{2 }(170)^{3 }− 5C_{3 }(170)^{2 }+ 5(170) − 1

A multiple of 100 + 5(170) – 1 = 100K + 849

∴ The last two digits are 49.

### Relation Between two Numbers

**Illustration: **Find the larger of 99^{50} + 100^{50} and 101^{50}

**Sol:**

101^{50 }= (100 + 1)^{50 }= 100^{50 }+ 50 . 100^{49 }+ 25 . 49 . 100^{48 }+ …

⇒ 99^{50 }= (100 − 1)^{50 }= 100^{50 }– 50 . 100^{49 }+ 25 . 49 . 100^{48 }− ….

⇒ 101^{50 }– 99^{50 }= 2[50 . 100^{49 }+ 25(49) (16) 100^{47 }+ …]

= 100^{50 }+ 50 . 49 . 16 . 100^{47 }+ … >100^{50}

∴ 101^{50 }– 99^{50 }> 100^{50}

⇒ 101^{50 }> 100^{50 }+ 99^{50}

### Divisibility Test

**Illustration: **Show that 11^{9 }+ 9^{11} is divisible by 10.

**Sol:**

11^{9 }+ 9^{11 }= (10 + 1)^{9 }+ (10 − 1)^{11}

= (9C_{0 }. 10^{9 }+ 9C_{1 }. 10^{8 }+ … 9C_{9}) + (11C_{0 }. 10^{11 }− 11C_{1 }. 10^{10 }+ … −11C_{11})

= 9C_{0 }. 10^{9 }+ 9C_{1 }. 10^{8 }+ … + 9C_{8 }. 10 + 1 + 10^{11 }− 11C_{1 }. 10^{10 }+ … + 11C_{10 }. 10−1

= 10[9C_{0 }. 10^{8 }+ 9C_{1 }. 10^{7 }+ … + 9C_{8 }+ 11C_{0 }. 10^{10 }− 11C_{1 }. 10^{9 }+ … + 11C_{10}]

= 10K, which is divisible by 10.

**Formulae:**

- The number of terms in the expansion of (x
_{1 }+ x_{2 }+ … x_{r})^{n}is (n + r − 1)C_{r – 1} - Sum of the coefficients of (ax + by)
^{n}is (a + b)^{n}

If f(x) = (a_{0 }+ a_{1}x + a_{2}x^{2 }+ …. + a_{m}x^{m})^{n} then

- (a) Sum of coefficients = f(1)
- (b) Sum of coefficients of even powers of x is: [f(1) + f(−1)] / 2
- (c) Sum of coefficients of odd powers of x is [f(1) − f(−1)] / 2

**Binomial Theorem for any Index**

Let n be a rational number and x be a real number such that | x | < 1 Then

**Proof:**

Let f(x) = (1 + x)^{n} = a_{0 }+ a_{1} x + a_{2} x^{2 }+ … +a_{r} x^{r }+ … (1)

f(0) = (1 + 0)^{n }= 1

Differentiating (1) w.r.t. x on both sides, we get

n(1 + x)^{n – 1}

= a_{1} + 2a_{2} x + 3a_{3} x^{3 }+ 4a_{4} x^{3 }+ … + ra_{r} x^{r – 1 }+ … (2)

Put x = 0, we get n = a_{1}

Differentiating (2) w.r.t. x on both sides, we get

n(n − 1)(1 + x)^{n – 2}

= 2a_{2 }+ 6a_{3} x + 12a_{4} x^{2 }+ … + r(r−1) a_{r} x^{r – 2 }+ … (3)

Put x = 0, we get a_{2 }= [n(n−1)] / 2!

Differentiating (3), w.r.t. x on both sides, we get

n(n − 1)(n − 2)(1 + x)^{n – 3 }= 6a_{3} + 24a_{4} x + … + r(r − 1)(r − 2) a_{r} x_{r – 3} + …

Put x = 0, we get a_{3 }= [n(n−1)(n−2)] / 3!

Similarly, we get a_{4 }= [n(n−1)(n−2)(n−3)] / r! and so on

∴ a_{r }= [n(n−1)(n−2)…(n−r+1)] / r!

Putting the values of a_{0}, a_{1}, a_{2}, a_{3}, …, a_{r} obtained in (1), we get

(1 + x)^{n }= 1 + nx + [{n(n−1)} / 2!] x^{2} + [{n(n − 1)(n − 2)} / 2!] x^{3 }+ … + [{n(n − 1)(n − 2) … (n – r + 1)}/ r!] x^{r }+ …

### Binomial theorem for Rational Index

The number of rational terms in the expression of (a^{1/l} + b^{1/k} )^{n} is [n / LCM of {l,k}] when none of and is a factor of and when at least one of and is a factor of is [n / LCM of {l,k}] + 1 where [.] is the greatest integer function.

**Illustration: **Find the number of irrational terms in (^{8}√5 + ^{6}√2)^{100}.

**Sol:**

T_{r + 1} = 100C_{r} (^{8}√5)^{100 – r }. (^{6}√2)^{r }= 100C_{r }. 5[(100 – r)/8] .2^{r/6}.

∴ r = 12,36,60,84

The number of rational terms = 4

Number of irrational terms = 101 – 4 = 97

### Binomial Theorem for Negative Index

**1. If rational number and -1 < x <1 then,**

- (1 − x)
^{-1 }= 1 + x + x^{2 }+ x^{3 }+ … + x^{r }+ … ∞ - (1 + x)
^{-1 }= 1 – x + x^{2 }– x^{3 }+ … (−1)^{r }x^{r}+ … ∞ - (1 − x)
^{-2}= 1 + 2x + 3x^{2}− 4x^{3}+ … + (r + 1)x^{r}+ … ∞ - (1 + x)
^{-2 }= 1 − 2x + 3x^{2 }− 4x^{3 }+ … + (−1)^{r }(r + 1)x^{r }+ … ∞

**2. Number of terms in (1 + x) ^{n} is**

- ‘n+1 when positive integer.
- Infinite when is not a positive integer & | x | < 1

**3. **First negative term in (1 + x)^{p/q} when 0 < x < 1, p, q are positive integers & ‘p’ is not a multiple of ‘q’ is T_{[p/q] }+ 3

## Multinomial Theorem

Using binomial theorem, we have

(x + a)^{n}

=^{n}∑_{r = 0}nC_{r} x^{n – r} a^{r}, n∈N

= ^{n}∑_{r = 0} [n! / (n − r)!r!] x^{n – r}a^{r}

= ^{n}∑_{r + s =n} [n! / r!s!] x^{s} a^{r}, where s = n – r.

This result can be generalized in the following form:

(x_{1} + x_{2} + … +x_{k})^{n}

= ∑_{r1 + r2 + …. + rk = n} [n! / r1!r2!…rk!] x_{1}^{r1} x_{2}^{r2} …x_{k}^{rk}

The general term in the above expansion is

[(n!) / (r_{1}! r_{2}! r_{3}! … r_{k}!)] x_{1}^{r1} x_{2}^{r2} x_{3}^{r3}… x_{k}^{rk}

The number of terms in the above expansion is equal to the number of non-negative integral solution of the equation.

r_{1}+r_{2 }+ … + r_{k }= n, because each solution of this equation gives a term in the above expansion. The number of such solutions is ^{n + k – 1}C_{k} −1.

**PARTICULAR CASES**

**Case-1:**

The above expansion has ^{n+3-1}C_{3-1} = ^{n + 2}C_{2} terms.

**Case-2: **

There are ^{n + 4 – 1}C_{4 – 1} = ^{n + 3}C_{3} terms in the above expansion.

**REMARK:** The greatest coefficient in the expansion of (x_{1 }+ x_{2 }+ … + x_{m})^{n} is [(n!) / (q!)^{m – r}{(q+1)!}^{r}], where q and r are the quotient and remainder, respectively when n is divided by m.

**Multinomial Expansions**

Consider the expansion of (x + y + z)^{10}. In the expansion, each term has different powers of x, y, and z and the sum of these powers is always 10.

One of the terms is λx^{2}y^{3}z^{5}. Now, the coefficient of this term is equal to the number of ways 2x′s, 3y′s, and 5z′s are arranged, i.e., 10! (2! 3! 5!). Thus,

(x+y+z)^{10 }= ∑(10!) / (P1! P2! P3!) x^{P1} y^{P2} z^{P3}

Where P1 + P2 + P3 = 10 and 0 ≤ P1, P2, P3 ≥ 10

In general,

(x_{1 }+ x_{2 }+ … x_{r})^{n }= ∑ (n!) / (P1! P2! … Pr!) x^{P1} x^{P2} … x^{Pr}

Where P1 + P2 + P3 + … + Pr = n and 0 ≤ P1, P2, … Pr ≥ n

**Number of Terms in the Expansion of **(x_{1} + x_{2} + … + x_{r})^{n}

From the general term of the above expansion, we can conclude that the number of terms is equal to the number of ways different powers can be distributed to x_{1}, x_{2}, x_{3 }…., x_{n} such that the sum of powers is always “n”.

Number of non-negative integral solutions of x_{1 }+ x_{2 }+ … + x_{r }= n is ^{n +r – 1}C_{r – 1}.

For example, number of terms in the expansion of (x + y + z)^{3} is ^{3 + 3 -1}C_{3 – 1 }= ^{5}C_{2 }= 10

As in the expansion, we have terms such as

As x^{0 }y^{0} z^{0}, x^{0} y^{1} z^{2}, x^{0} y^{2} z^{1}, x^{0} y^{3} z^{0}, x^{1} y^{0} z^{2}, x^{1} y^{1} z^{1}, x^{1} y^{2} z^{0}, x^{2} y^{0} z^{1}, x^{2} y^{1} z^{0}, x^{3} y^{0} z^{0}.

Number of terms in (x + y + z)^{n} is ^{n + 3 – 1}C_{3 – 1 }= ^{n + 2}C_{2}.

Number of terms in (x + y + z + w)^{n} is ^{n + 4 – 1}C_{4 – 1} = ^{n + 3}C_{3} and so on.

### Binomial Theorem IIT JEE Video Lesson

### Binomial Coefficients with G.P. or A.P.

### Binomial Theorem – Top 12 Most Important and Expected Questions

## Problems on Binomial Theorem

**Question 1: If the third term in the binomial expansion of equals 2560, find x.**

**Solution:**

⇒ (log_{2}^{x})^{2} = 4

⇒ log_{2}^{x} = 2 or -2

⇒ x = 4 or 1/4.

**Question 2: Find the positive value of λ for which the coefficient of x ^{2 }in the expression x^{2}[**

**√**x

**+ (λ/x**

^{2})]^{10}is 720.**Solution:**

⇒ x^{2 }[^{10}C_{r} . (√x)^{10-r} . (λ/x^{2})^{r}] = x^{2} [^{10}C_{r }. λ^{r} . x^{(10-r)/2} . x^{-2r}]

= x^{2} [^{10}C_{r }. λ^{r} . x^{(10-5r)/2}]

Therefore, r = 2

Hence, ^{10}C_{2} . λ^{2} = 720

⇒ λ^{2} = 16

⇒ λ = ±4.

**Question 3: The sum of the real values of x for which the middle term in the binomial expansion of (x ^{3}/3 + 3/x)^{8} equals 5670 is?**

**Solution:**

T_{5} = ^{8}C_{4} × (x^{12}/81) × (81/x^{4}) = 5670

⇒ 70 x^{8} = 5670

⇒ x = ± √3.

**Question 4: Let (x + 10) ^{50 }+ (x – 10)^{50} = a_{0} +a_{1}x + a_{2} x_{2} + . . . . . + a_{50} x^{50} for all x ∈R, then a_{2}/a_{0} is equal to?**

**Solution:**

⇒ (x + 10)^{50 }+ (x – 10)^{50}:

a_{2} = 2 × ^{50}C_{2} × 10^{48}

a_{0} = 2 × 10^{50}

⇒ a_{2}/a_{0} = ^{50}C_{2}/10^{2} = 12.25.

**Question 5: Find the coefficient of x ^{9} in the expansion of (1 + x) (1 + x^{2} ) (1 + x^{3}) . . . . . . (1 + x^{100}).**

**Solution: **

x^{9} can be formed in 8 ways.

i.e., x^{9 }x^{1+8 }x^{2+7 }x^{3+6 }x^{4+5}, x^{1+3+5}, x^{2+3+4}

∴ The coefficient of x^{9 }= 1 + 1 + 1 + . . . . + 8 times = 8.

**Question 6: The coefficients of three consecutive terms of (1 + x) ^{n+5 }are in the ratio 5:10:14, find n.**

**Solution:**

Let T_{r-1}, T_{r}, T_{r+1 }are three consecutive terms of (1 + x)^{n+5}

⇒ T_{r-1} = (n+5) C_{r-2} . x^{r-2}

⇒ T_{r} = (n+5) C_{r-1} . x^{r-1}

⇒ T_{r+1} = (n+5) C_{r} . x^{r}

Given

(n+5) C_{r-2} : (n+5) C_{r-1 }: (n+5) C_{r} = 5 : 10 : 14

Therefore, [(n+5) C_{r-2}]/5= [(n+5) C_{r-1}]/10 = (n+5) C_{r}/14

Comparing first two results we have n – 3r = -9 . . . . . . (1)

Comparing last two results we have 5n – 12r = -30 . . . . . . (2)

From equation (1) and (2) n = 6.

**Question 7: The digit in the units place of the number 183! + 3 ^{183}**

**.**

**Solution: **

⇒ 3^{183 }= (3^{4})^{45}.3^{3}

⇒ unit digit = 7 and 183! ends with 0

∴ Units digit of 183! + 3^{183 }is 7.

**Question 8: Find the total number of terms in the expansion of **(x + a)^{100 }+ (x – a)^{100}.

**Solution:**

**⇒ (x + a) ^{100 }+ (x – a)^{100 }**= 2[

^{100}C

_{0}x

^{100}.

^{100}C

_{2 }x

^{98}. a

^{2}+ . . . . . . +

^{100}C

_{100 }a

^{100}]

∴ Total Terms = 51.

**Question 9: Find the coefficient of t ^{4} in the expansion of [(1-t^{6})/(1 – t)].**

**Solution: **

**⇒ [(1-t ^{6})/(1 – t)] = **(1 – t

^{18}– 3t

^{6}+ 3t

^{12}) (1 – t)

^{-3}

Coefficient of t in (1 – t)^{-3 }= 3 + 4 – 1

C_{4} = ^{6}C_{2} =15

The Coefficient of x^{r} in **(1 – x) ^{-n }**= (r + n – 1) C

_{r}

**Question 10:** Find the ratio of the 5^{th} term from the beginning to the 5^{th} term from the end in the binomial expansion of [2^{1/3} + 1/{2.(3)^{1/3}}]^{10}**. **

**Solution: **

**Question 11: Find the coefficient of **a^{3}b^{2}c^{4}d **in the expansion of (a-b-c+d) ^{10}.**

**Solution:**

Expand** (a – b – c + d) ^{10}** using multinomial theorem and by using coefficient property we can obtain the required result.

Using multinomial theorem, we have

We want to get coefficient of a^{3}b^{2}c^{4}d this implies that r_{1} = 3, r_{2} = 2, r_{3} = 4, r_{4} = 1,

∴ The coefficient of a^{3}b^{2}c^{4}d is [(10)!/(3!.2!.4)] (-1)^{2} (-1)^{-4} = 12600.

**Question 12: Find the coefficient of in the expansion of (1 + x + x ^{2 }+x^{3})^{11}.**

**Solution:**

By expanding given equation using expansion formula we can get the coefficient x^{4}

i.e. 1 + x + x^{2 }+ x^{3 }= (1 + x) + x^{2 }(1 + x) = (1 + x) (1 + x^{2})

⇒ (1 + x + x^{2 }+ x^{3}) x^{11 }= (1+x)^{11 }(1+x^{2})^{11}

= 1+ ^{11}C_{1 }x^{2 }+ ^{11}C_{2 }x^{2 }+ ^{11}C_{3 }x^{3 }+ ^{11}C_{4 }x^{4 }. . . . . . .

= 1 + ^{11}C_{1 }x^{2 }+ ^{11}C_{2 }x^{4 }+ . . . . . .

To find term in from the product of two brackets on the right-hand-side, consider the following products terms as

= 1 × ^{11}C_{2 }x^{4 }+ ^{11}C_{2 }x^{2} × ^{11}C_{1 }x^{2 }+ ^{11}C_{4 }x^{4}

= ^{11}C_{2 }+ ^{11}C_{2 }× ^{11}C_{1 }+ ^{11}C_{4 }] x^{4}

⇒ [55 + 605 + 330] x^{4} = 990x^{4}

∴ The coefficient of x^{4} is 990.

**Question 13: Find the number of terms free from the radical sign in the expansion of (√5 + ^{4}√n)^{100}.**

**Solution:**

T_{r+1} = ^{100}C_{r} . 5^{(100 – r)/2} n^{r/4}

Where r = 0, 1, 2, . . . . . . , 100

r must be 0, 4, 8, … 100

Number of rational terms = 26

**Question 14: Find the degree of the polynomial [x + {√(3 ^{(3-1)})}^{1/2}]^{5} + [x + {√(3^{(3-1)})}^{1/2}]^{5}.**

**Solution:**

[x + { √(3^{(3-1)}) }^{1/2 }]^{5}:

= 2 [^{5}C_{0} x^{5} + ^{5}C_{2} x^{5} (x^{3} – 1) + ^{5}C_{4} . x . (x^{3} – 1)^{2}]

Therefore, the highest power = 7.

**Question 15: Find the last three digits of 27 ^{26}**

**.**

**Solution: **

By reducing 27^{26} into the form (730 – 1)^{n} and using simple binomial expansion we will get required digits.

We have **27 ^{2 }= 729**

Now 27^{26} = (729)^{13} = (730 – 1)^{13}

= ^{13}C_{0 }(730)^{13} – ^{13}C_{1 }(730)^{12} + ^{13}C_{2 }(730)^{11} – . . . . . – ^{13}C_{10 }(730)^{3} + ^{13}C_{11}(730)^{2} – ^{13}C_{12 }(730) + 1

= 1000m + [(13 × 12)]/2] × (14)^{2} – (13) × (730) + 1

Where ‘m’ is a positive integer

= 1000m + 15288 – 9490 = 1000m + 5799

Thus, the last three digits of 17^{256} are 799.

## Binomial Theorem JEE Solution

## Binomial Theorem – Important Topics

## Binomial Theorem – Important Questions

## Frequently Asked Questions

### Give the Binomial Theorem formula.

We use the Binomial theorem to find the expansion of the algebraic terms of the form(x + y)^{n}. The formula is (x + y)^{n} = Σ_{r=0}^{n} ^{n}C_{r} x^{n – r} · y^{r}.

### What is the general term in a Binomial expansion?

The general term of a binomial expansion is T_{r+1} = ^{n}C_{r} x^{n-r} y^{r}.

### What is the number of terms in the expansion of (x + a)^{n} + (x-a)^{n} ?

The number of terms in the expansion of (x + a)^{n} + (x-a)^{n} are (n+2)/2 if n is even, or (n+1)/2 if n is odd.

### List two applications of the Binomial Theorem.

In Mathematics, Binomial theorem is used to find the remainder and also to find the digits of a number.