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

Introduction to the Binomial Theorem

The Binomial Theorem is a powerful tool of expansion which is used to expand an expression that has been raised to any finite power. It has applications in Algebra, probability, etc.

Binomial Expression: An algebraic expression that consists of two terms that are not the same is known as a binomial expression. For example: a + b, a3 + b3, etc.

Binomial Theorem: Let $n\in \mathbb{N}$, $x,y\in \mathbb{R}$ then

$(x+y{)}^n=\sum _{r=0}^n(\genfrac{}{}{0ex}{}{n}{r})x^{n-r}{y}^r$ where,

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

Given:

I need help

Sol:

What kind of help do you need?

(√2 + 1)⁵ + (√2 - 1)⁵

Given:

This is a sentence.

Sol:

This is a sentence.

We have

$(x+y{)}^5+(x-y{)}^5=2(\genfrac{}{}{0ex}{}{5}{0})x^5+(\genfrac{}{}{0ex}{}{5}{2})x^3{y}^2+(\genfrac{}{}{0ex}{}{5}{4})x^1{y}^4$

2x^5 + 10x^3y^2 + 5xy^4

Now $(\sqrt{2}+1{)}^5+(\sqrt{2}-1{)}^5=2[(\sqrt{2}{)}^5+10(\sqrt{2}{)}^3(1{)}^2+5(\sqrt{2})(1{)}^4]$

58<sup>√2</sup>

Binomial Expansion

Important Points to Remember:

The total number of terms in the expansion of $(x+y{)}^n$ are $(n+1)$

The sum of the exponents of x and y is always equal to n.

The terms nC0, nC1, nC2, ….., nCn are known as binomial coefficients and can also be represented by C0, C1, C2, ….., Cn.

The binomial coefficients which are equidistant from the beginning and from the ending are equal, i.e.

nC₀ = nC_n, nC₁ = nC_n-1, nC₂ = nC_n-2, … etc.

We can use Pascal’s Triangle to find binomial coefficients.

Using Pascal’s Triangle to Find Binomial Coefficients

More Useful Expansions:

$(x + y)^n + (x - y)^n = 2[C_0 x^n + C_2 x^{n-1} y^2 + C_4 x^{n-4} y^4 + \cdots]$

$(x+y{)}^n-(x-y{)}^n=2[C_1{x}^{n-1}y+C_3{x}^{n-3}{y}^3+C_5{x}^{n-5}{y}^5+\dots ]$

(1 + x)ⁿ = nΣ_r=0 nCr . x^r = [C₀ + C₁ x + C₂ x² + … C_n xⁿ]

(1 + x)ⁿ + (1 - x)ⁿ = 2[C₀ + C₂x² + C₄x⁴ + …]

(1+x)ⁿ - (1-x)ⁿ = 2[C₁x + C₃x³ + C₅x⁵ + …]

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

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

The Relationship between the Number of Terms and R-F Factor

Properties of the 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:

C0 + C1 + C2 + … + Cn = 2n

C₀ + C₂ + C₄ + … = C₁ + C₃ + C₅ + … = 2n-1

C₀ - C₁ + C₂ - C₃ + … + (-1)ⁿ C_n = 0

nC₁ + 2nC₂ + 3nC₃ + … + nnC_n = n(2ⁿ - 1)

C₁ - 2C₂ + 3C₃ - 4C₄ + ... + (-1)ⁿ⁻¹Cⁿ = 0 for n > 1

C₀₂ + C₁₂ + C₂₂ + …C_n² = [(2n)!/ (n!)²]

Answer: If is equal to $(1+x{)}^{15}=a_0+a_{1}x+\cdots +a_{15}{x}^{15}$, then find the value of $x$.

Given Text

Welcome to Our Site

Sol:

Welcome to Our Site :smiley:

C0/C0 + C1/C0 + 2C2/C1 + 3C3/C2 + ... + 15C15/C14

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

[Properties of Binomial Coefficients Video Lesson]

Terms in the Binomial Expansion

The different terms in the binomial expansion covered here are:

• Middle term
• General term

General Term

Middle Term

Independent Term

Determining a Particular Term

Numerically Most Significant Term

Ratio of Consecutive Terms/Coefficients

Binomial Coefficients

We have $(x+y{)}^n=(\genfrac{}{}{0ex}{}{n}{0})x^n+(\genfrac{}{}{0ex}{}{n}{1})x^{n-1}y+(\genfrac{}{}{0ex}{}{n}{2})x^{n-2}{y}^2+\dots +(\genfrac{}{}{0ex}{}{n}{n})y^n$

General Term = $T_r+1=(\genfrac{}{}{0ex}{}{n}{r})\cdot x^{n-r}\cdot y^r$

(1 + x)^n = nC_rx^r

The (n - r + 2)th term from the end of the binomial expansion of (x + y)n is the rth term.

Answer: The number of terms in $(1+2x+x^2{)}^{50}$ is 51.

Given:

I like to go to the beach

Sol:

I enjoy going to the beach

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

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

Answer: -1/x¹²

Given:

This is a heading

Sol:

This is a heading

T10 - 4 + 2 = T8 = 10C7 (2x)<sup>3</sup> (-1/x<sup>2</sup>)<sup>7</sup> = -960x<sup>-11</sup>

$(x+y{)}^n$

If n is even, then the $(\frac{n}{2}+1{)}^{\text{th}}$ Term is the middle Term.

If n is odd, then the [(n+1)/2]th and [(n+3)/2]th terms are the middle terms.

Illustration: Find the middle term of $(1-3x+3x^2-x^3{)}^{2n}$

Given:

This is a heading

Sol:

This is a heading

$(1-3x+3x^2-x^3{)}^{2n}=[(1-x{)}^3{]}^{2n}=(1-x{)}^{6n}$

Middle Term = $$\frac{6n}{2}+1\text{term }=6nC3n(-x{)}^{3}n$$

Determining a Particular Term:

The coefficient of $x^m$ in the expansion of $(a{x}^p+b/x^q{)}^n$ is the coefficient of $T^r+1$ where $r=\frac{np-m}{p+q}$.

In the expansion of $(x+a{)}^n$, $\frac{T_{r+1}}{T_r}=\frac{n-r+1}{r}\frac{a}{x}$

General and Middle Terms of a Binomial Expansion

Dependent Term

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

Tr+1 = ${n \choose r} \frac{a^n-r}{b^r}, \text{where } r = \left\lfloor \frac{np}{p+q} \right\rfloor

Answer: 6

This is a **bold** statement.

Sol: This is a bold statement.

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

6C3 = 20

Answer: Find the independent term in the expansion of: (x + y)^2

Given:

This is a Heading

Sol:

This is a Heading

(x^1/3 + 1 - 1 - x^-1/2)¹⁰ = (x^1/3 - x^-1/2)¹⁰

r = \frac{10 \cdot \frac{1}{3}}{\frac{1}{3} + \frac{1}{2}} = 4

∴ T5 = 10C4 = 210

Numerically Greatest Term in the Expansion of $(1+x{)}^n$:

$x^n$

If $$\frac{(n+1)|x|}{|x|+1}=P,$$ where P is a positive integer, then the Pth and (P+1)th terms are numerically the greatest terms in the expansion of $(1+x{)}^n$.

If $$\frac{|n+1||x|}{|x|+1}=P+F,$$ where $P$ is a positive integer and $0<F<1$, then the $(P+1)$th term is numerically the greatest term in the expansion of $(1+x{)}^n$.

Answer: The numerically greatest term in (1-3x)¹⁰ when x = (1/2) is 1.

Given:

This is a Heading

Sol:

This is a Heading

[(n + 1) \|\alpha\|] / \|\alpha\| + 1 = \frac{11 \times 3/2}{3/2+1} = \frac{33}{5} = 6.6

Therefore, T7 is the numerically highest term.

T^6 + 1 = \binom{10}{6} \quad (-3x)^6 = \binom{10}{6} \quad \left(\frac{3}{2}\right)^6

Ratio of Consecutive Terms/Coefficients:

Coefficients of $x^r$ and $x^r+1$ are $(\genfrac{}{}{0ex}{}{n}{r})-1$ and $(\genfrac{}{}{0ex}{}{n}{r})$ respectively.

(nCr / nCr-1) = (n-r+1)/r

Question: What is the value of n if the coefficients of three consecutive terms in the expansion of (1+x)n are in the ratio 1:7:42?

Given:

This is a heading

Sol:

This is a heading

Let $(r-1)$th, $r$th, and $(r+1)$th be the three consecutive terms.

The ratio is 1:7:42.

Now $$\frac{(\genfrac{}{}{0ex}{}{n}{r})-2}{(\genfrac{}{}{0ex}{}{n}{r})-1}=\frac{1}{7}$$

(nCr-2 / nCr-1) = (1/7) ⇒ [(r-1)/(n-r+2)] = (1/7) ⇒ n-8r+9 = 0 → (1)

⇒ [(r-1)/(n-r+2)] = (1/7) ⇒ n-8r+9 = 0 → (1) he said.

He said, “And.”

r = (7n - 42)/(n - 1) ⇒ [(7n - 42)/(n - 1)]/(n) = (1/6) ⇒ 7n - 42 = 6n ⇒ n = 42/7 → (2)

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

Applications of Binomial Theorem

1. Computing powers of binomials
2. Solving polynomial equations
3. Computing probabilities in probability theory
4. Finding the coefficients of binomial expansions

The most common applications of the Binomial Theorem are:

• Finding the remainder
• Finding digits of a number
• Other applications in Mathematics

Calculating the Remainder Using the Binomial Theorem

Answer: The remainder when 7103 is divided by 25 is 8.

#I love to go to the movies

Sol: I absolutely adore going to the movies!

(7103 / 25) = [7(49)51 / 25] = [7(50 - 1)51 / 25]

(7103 / 25) = 284.12 = [7(50 - 1)51 / 25] = [74951 / 25] = 299.84

[175K - 7]/25

\frac{25(7K - 1) + 18}{25}

Therefore, the remainder is 18.

Answer: Find K if the fractional part of the number (2403 / 15) is (K/15).

Given:

Welcome to Our Website

Sol:

Welcome to Our Website :smiley:

(2403 / 15) = 160.2

8/15 (15 + 1)100 = 8/15 (15λ + 1) = 8λ + 8/15

The fractional part of 8λ is 8/15, since 8λ is an integer.

K = 8.

Calculating the Digits of a Number

Answer: The last two digits of (13)10 are 13.

Given Text:

This is a Header

Sol:

This is a Header

(13)_{10} = (169)_5 = (170 - 1)_5

5C₀ (170)⁵ - 5C₁ (170)⁴ + 5C₂ (170)³ - 5C₃ (170)² + 5C₄ (170) - 5C₅

5C₀ (170)⁵ - 5C₁ (170)⁴ + 5C₂ (170)³ - 5C₃ (170)² + 5(170) - 1

100K + 848 = 100 × (5 + 17) - 1

Therefore, the last two digits are 49.

Relationship Between Two Numbers

Answer: Which is larger, 9950 + 10050 or 10150?

Given:

What is your favorite color?

Sol: My favorite color is blue.

10150 = (100 + 1)50 = 10050 + 50

10150 = 10050 + 50 * 10049 + 25 * 49 * 10048 + …

⇒ 9950 = (100 - 1)50 = 10050 - 50 \ 10049 + 25 \ 49 \ 10048 - …

⇒ 10150 - 9950 = 250. 10049 + 25(49) = 16(10047) + ...

10050 + 50 + 49 + 16 + 10047 + ... > 10050

10150 - 9950 > 10050

10150 > 20000

Divisibility Test

Solution: 119 + 911 = 1030, which is divisible by 10.

Given:

This is a heading

Sol:

This is a heading

(10 + 1)9 + (10 - 1)11 = 119 + 911

=(9C_0 \cdot 10^9 + 9C_1 \cdot 10^8 + \ldots + 9C_9) + (11C_0 \cdot 10^{11} - 11C_1 \cdot 10^{10} + \ldots - 11C_{11})

9C0 . 109 + 9C1 . 108 + 9C2 . 107 + 9C3 . 106 + 9C4 . 105 + 9C5 . 104 + 9C6 . 103 + 9C7 . 102 + 9C8 . 10 + 1 + 11C1 . 1010 + 11C2 . 109 + 11C3 . 108 + 11C4 . 107 + 11C5 . 106 + 11C6 . 105 + 11C7 . 104 + 11C8 . 103 + 11C9 . 102 + 11C10 . 10 - 1

10^9C_0 \cdot 10^8 + 9C_1 \cdot 10^7 + \ldots + 9C_8 + 11C_0 \cdot 10^{10} - 11C_1 \cdot 10^9 + \ldots + 11C_{10}

10K is divisible by 10.

Formulae:

The number of terms in the expansion of $(x_1+x_2+\cdots +x_r{)}^n$ is $(\genfrac{}{}{0ex}{}{n+r-1}{r-1})$.

The sum of the coefficients of $(ax+by{)}^n$ is $(a+b{)}^n$

If $$f(x)=(a_0+a_{1}x+a_2{x}^2+\dots +a_m{x}^m{)}^n$$ then

The sum of coefficients = f(1)

Sum of coefficients of even powers of x is: [f(1) + f(-1)] / 2

The sum of coefficients of odd powers of x is [f(1) - f(-1)]/2

The Binomial Theorem for Any Index

Let $n$ be a rational number and $x$ be a real number such that $|x|<1$, then

Proof:

Let $f(x)=(1+x{)}^n=a_0+a_{1}x+a_2{x}^2+\cdots +a_r{x}^r+\dots$ (1)

f(0) = 1ⁿ = 1

Differentiating both sides of (1) with respect to $x$, we get

n(1 + x)ⁿ - 1

= $\sum _{i=1}^r{a}_i{x}^i-1$

If we set x = 0, then n = a1.

Differentiating both sides of (2) with respect to $x$, we get

$n(n-1)(1+x{)}^n-2$

$2a^2+6a^{3}x+12a^4{x}^2+\dots +r(r-1)ar{x}^r-2+\dots (3)$

If x = 0, then we get $$a_2=\frac{n(n-1)}{2}!$$

Differentiating both sides of (3) with respect to x, we get

n(n - 1)(n - 2)(1 + x)^{n - 3} = 6a³ + 24a⁴x + … + r(r - 1)(r - 2)ar^{xr - 3} + …

If x = 0, then a3 = [n(n−1)(n−2)] / 3!

Similarly, we get $$a_4=\frac{n(n-1)(n-2)(n-3)}{r!}$$ and so on.

∴ ar = [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

$(1+x{)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{2!}{x}^3+\cdots +\frac{n(n-1)(n-2)\dots (n-r+1)}{r!}{x}^r+\dots$

The Binomial Theorem for a Rational Index

The number of rational terms in the expression of $${(\frac{a_1}{l}+\frac{b_1}{k})}^n$$ is $$\left[\frac{n}{\text{LCM of }l,k}\right]$$ when none of $l$ and $k$ is a factor of $n$, and when at least one of $l$ and $k$ is a factor of $n$ is $$\left[\frac{n}{\text{LCM of }l,k}\right]+1$$ where $[\cdot ]$ is the greatest integer function.

Answer: Find the number of irrational terms in $(8\sqrt{5}+6\sqrt{2}{)}^{100}$.

Given:

This is bold text

Sol:

This is bold text

Tr + 1 = 100Cr $\times$ $(8\sqrt{5}{)}^{100-r}\times (6\sqrt{2}{)}^r$ = 100Cr $\times$ $5\times \left[\frac{(100-r)}{8}\right]\times \left[\frac{2r}{6}\right]$

r = 12, 36, 60, 84

The number of rational terms = 4

Number of irrational terms = 101 - 4 = 97

Binomial Theorem for Negative Exponents

If a rational number is such that -1 < x < 1, then

$\frac{1}{1-x}=1+x+x^2+x^3+\dots +x^r+\dots \mathrm{\infty }$

$(1+x{)}^{-1}=1-x+x^2-x^3+\cdots (-1{)}^r{x}^r+\cdots \mathrm{\infty }$

$(1-x{)}^{-2}=1+2x+3x^2-4x^3+\cdots +(r+1)x^r+\dots \mathrm{\infty }$

$(1+x{)}^{-2}=1-2x+3x^2-4x^3+\dots +(-1{)}^r(r+1)x^r+\dots \mathrm{\infty }$

![Binomial Theorem]()

The number of terms in $(1+x{)}^n$ is $n+1$

’n+1 when n is a positive integer.’

Infinite is not a positive integer when |x| < 1

3. The first negative term in $(1+x{)}^{\frac{p}{q}}$ when $0<x<1$, $p,q$ are positive integers and $p$ is not a multiple of $q$ is $T_{\frac{p}{q}}+3$.

Multinomial Theorem

Using the binomial theorem, we can

$(x+a{)}^n$

n\sum_{r=0}^{n}\text{C}_{r}x^{n-r}a^{r}, \quad n \in \mathbb{N}

$$\sum _{r=0}^n\frac{n!}{(n-r)!r!}{x}^{n-r}{a}^r$$

n∑r + s = n [n! / (r!)(s!)] xs^r, where s = n - r.

This result can be generalized as follows:

(x1 + x2 + … + xk)<sup>n</sup>

$$\sum _{i=1}^k{r}_i=n[\frac{n!}{r_1!r_2!\cdots r_k!}{x}_1^{r_1}{x}_2^{r_2}\cdots x_k^{r_k}]$$

The general term in the above expansion is $a_n$

[\frac{n!}{r_1! r_2! r_3! \dots r_k!}] \times x_1^{r_1} x_2^{r_2} x_3^{r_3} \dots x_k^{r_k}]

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

The number of solutions to the equation $$r_1+r_2+\cdots +r_k=n$$ is $$(\genfrac{}{}{0ex}{}{n+k-1}{k-1})$$, because each solution of this equation gives a term in the above expansion.

Particular Cases

Case-1:

The above expansion has $$n+3-1(\genfrac{}{}{0ex}{}{3-1}{2})=n+2(\genfrac{}{}{0ex}{}{2}{2})$$ terms.

Case-2:

There are $n+3C3$ terms in the above expansion.

REMARK: The greatest coefficient in the expansion of $(x_1+x_2+\cdots +x_m{)}^n$ is $\frac{n!}{q!m-r(q+1){!}^r}$, where $q$ and $r$ are the quotient and remainder, respectively when $n$ is divided by $m$.

Multinomial Expansions

The expansion of $(x+y+z{)}^{10}$ will contain terms with different powers of $x$, $y$, and $z$ such that the sum of these powers always equals 10.

The coefficient of the term λx2y3z5 is 10! (2! 3! 5!) since it is equal to the number of ways 2x′s, 3y′s, and 5z′s are arranged.

$(x^P1y^P2z^P3)*10 = \frac{\sum(10!)}{P1! P2! P3!}$

0 ≤ P1 + P2 + P3 ≤ 10

In general,

$(x_1+x_2+\dots x_r{)}^n=\sum \frac{n!}{P_1!P_2!\dots P_r!}{x}_{P_1}{x}_{P_2}\dots x_{P_r}$

P1 + P2 + P3 + … + Pr = n, where 0 ≤ P1, P2, … Pr ≤ n

Number of Terms in the Expansion of $(x_1+x_2+\cdots +x_r{)}^n$

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 $n$.

The number of non-negative integral solutions of $$x_1+x_2+\cdots +x_r=n$$ is $$(\genfrac{}{}{0ex}{}{n+r-1}{r-1})$$

The number of terms in the expansion of $(x+y+z{)}^3$ is $(\genfrac{}{}{0ex}{}{3}{3})-(\genfrac{}{}{0ex}{}{3}{1})+1=(\genfrac{}{}{0ex}{}{5}{2})=10$

As seen in the expansion, terms such as

x0 y0 z0,
x0 y1 z2,
x0 y2 z1,
x0 y3 z0,
x1 y0 z2,
x1 y1 z1,
x1 y2 z0,
x2 y0 z1,
x2 y1 z0,
x3 y0 z0.

The number of terms in $(x+y+z{)}^n$ is $(\genfrac{}{}{0ex}{}{n+3-1}{3-1})-1=(\genfrac{}{}{0ex}{}{n+2}{2})$.

The number of terms in $(x+y+z+w{)}^n$ is $n+4-(\genfrac{}{}{0ex}{}{4}{1})-1=n+3(\genfrac{}{}{0ex}{}{3}{1})$ and so on.

Binomial Theorem IIT JEE Video Lesson

Binomial Coefficients with Geometric Progression or Arithmetic Progression

Top 12 Most Important and Expected Questions on the Binomial Theorem

Problems on Binomial Theorem

Question 1: If the third term in the binomial expansion of equals 2560, what is the value of x?

Given:

This is a heading

Solution:

This is a heading

(log_2x)^2 = 4

log_2x = 2 \text{ or } -2

x = 4 or 0.25

Question 2: Find the value of λ such that the coefficient of x2 in the expression x2[√x + (λ/x2)]10 is 720.

Given:

This is a heading

Solution:

This is a heading

x2 [10Cr . λr . (x/2)(10-r) . (x²)-2r] = x2 [10Cr . (√x)10-r . (λ/x2)r]

$x^2 = \frac{10Cr \cdot \lambda r \cdot x(10-5r)}{2}$

Therefore, r = 2

Hence, $$(\genfrac{}{}{0ex}{}{10}{2}){\lambda }^2=720$$

λ² = 16

λ = ±4

Question 3: What is 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?

Given:

This is a heading

Solution:

This is a heading

T5 = 8C<sup>4</sup> × (x<sup>12</sup>/81) × (81/x<sup>4</sup>) = 5670

70 x 8 = 560

x = ±√3

Question 4: If $(x+10{)}^{50}+(x-10{)}^{50}=a_0+a_{1}x+a_2{x}^2+\cdots +a_{50}{x}^{50}$ for all $x\in \mathbb{R}$, then what is $\frac{a_2}{a_0}$?

Given:

This is a heading

Solution:

This is a heading

(x + 10) \* 50 + (x - 10) \* 50

a2 = 2 × 50C2 × 1048

a0 = 2 × 10⁵⁰

a2/a0 = \frac{50C2}{102} = 12.25

Answer 5: The coefficient of x9 in the expansion of (1 + x) (1 + x2 ) (1 + x3) . . . . . . (1 + x100) is 100.

Given:

This is a heading

Solution:

This is a heading

x9 can be formed in 8 ways.

i.e., x₉ x₁+8 x₂+7 x₃+6 x₄+5, x₁+3+5, x₂+3+4

The coefficient of x^9 = 1 + 1 + 1 + ... + 8 = 8 \times 8.

Question 6: If the coefficients of three consecutive terms of $(1+x{)}^n+5$ are in the ratio 5:10:14, what is the value of $n$?

Given:

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Solution:

This is a header

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

Tr-1 = (n+5) x Cr-2

Tr = (n+5)Cr⁻¹ xr⁻¹

Tr+1 = (n+5)Cr * xr

Given

This is a statement.

Given This is a statement.

(n+5) Cr-2 : 5

(n+5) Cr-1 : 10

(n+5) Cr : 14

Therefore, $$\frac{(n+5)Cr-2}{5}=\frac{(n+5)Cr-1}{10}=\frac{(n+5)Cr}{14}$$

Comparing the first two results, we have $n-3r=-9$ (1)

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

From equations 1 and 2, we get $n=6$.

Answer 7: The digit in the units place of the number 183! + 3183 is 4.

Given:

This is a heading

Solution:

This is a heading

3183 = (34) x 45.33

The unit digit of 183! is 7 and it ends with 0.

The unit digit of the sum of 183! and 3183 is 7.

Answer: 200

Given: This is a sentence.

Solution: This is a sentence.

(x + a)¹⁰⁰ + (x - a)¹⁰⁰ = 2¹⁰⁰C₀x¹⁰⁰ + 2¹⁰⁰C₂x⁹⁸a² + … + 2¹⁰⁰C₁₀₀a¹⁰⁰

Total Terms = 51.

Answer 9: The coefficient of t⁴ in the expansion of [(1-t⁶)/(1 - t)] is -t⁴.

Given:

Welcome to the store!

Solution:

Welcome to the store! :smiley:

⇒ [(1-t6)/(1 - t)] = (1 - t18 - 3t6 + 3t12)/(1 - t)

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

C4 = 6C2 = 15

The Coefficient of x^r in (1 - x)^-n = (r + n - 1) C_r

Question 10: Find the ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of [21/3 + 1/{2.(3)1/3}]¹⁰?

Given:

This is a heading

Solution:

This is a heading

Answer 11: The coefficient of a³b²c⁴d in the expansion of (a-b-c+d)¹⁰ is -10240.

Given:

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Solution:

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Expand (a¹⁰ – b¹⁰ - c¹⁰ + d¹⁰) using the multinomial theorem and the coefficient property to obtain the required result.

Using the multinomial theorem, we have

![Binomial Theorem Problems]()

We want to get the coefficient of a^3b^2c^4d, which implies that r1 = 3, r2 = 2, r3 = 4, and r4 = 1.

∴ The coefficient of a³b²c⁴d is [(10)!/(3!².4!)] (-1)² (-1)^-4 = 12600.

Answer 12: The coefficient of $x^9$ in the expansion of $(1+x+x^2+x^3{)}^{11}$ is 11,279.

Given:

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Solution:

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By applying the expansion formula to the given equation, we can obtain the coefficient of x⁴.

(1 + x) + x^2(1 + x) = (1 + x) (1 + x^2)

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

1 + \binom{11}{1}x^2 + \binom{11}{2}x^2 + \binom{11}{3}x^3 + \binom{11}{4}x^4 + \cdots

1 + 11C1x2 + 11C2x4 + 11C3x6 + 11C4x8 + 11C5x10 + 11C6x12

Consider the following products terms as a way to find the term in the product of two brackets on the right-hand-side.

11C2x4 + 11C2x2 × 11C1x2 + 11C4x4 = 1

[11C2 + 11C2 × 11C1 + 11C4] x 4

⇒ [55 + 605 + 330] x 4 = 2790

The coefficient of x^4 is 990.

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

Given:

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Solution:

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Tr+1 = $\frac{100Cr\cdot 5(100-r)}{2nr/4}$

Where $r=0,1,2,\dots ,100$

r must be 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, or 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.

Given:

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Solution:

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[x + (3√2)⁵⁄₂]⁵

5C₀ x⁵ + 5C₂ x⁵ (x³ - 1) + 5C₄ x (x³ - 1)²

Therefore, the highest power = 7.

Answer: 726

Given:

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Solution:

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By expressing 2726 as $(730-1)n$ and using the binomial theorem, we can obtain the desired digits.

We have 729 = 272

Now 2726 = 729<sup>13</sup> = (730 - 1)<sup>13</sup>

13C₀(730)¹³ - 13C₁(730)¹² + 13C₂(730)¹¹ - ... - 13C¹⁰(730)³ + 13C¹¹(730)² - 13C¹²(730) + 1

1000m + [(13 × 12)/2] × (14)^2 - (13) × (730) + 1

Where m is a positive integer.

1000m + 15288 - 9490 = 5799

Thus, the last three digits of 17256 are 256.

Binomial Theorem JEE Solutions

Binomial Theorem: An Important Topic

Binomial Theorem - Important Questions

Frequently Asked Questions

The Binomial Theorem Formula is: $(x+y{)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})x^k{y}^{n-k}$

$(x+y{)}^n=\sum _{r=0}^n(\genfrac{}{}{0ex}{}{n}{r})x^{n-r}{y}^r$

What is the General Term in a Binomial Expansion?

The general term of a binomial expansion is T_r+1 = n_Cr x^n-r y^r.

The number of terms in the expansion of $(x+a{)}^n+(x-a{)}^n$ is $2n$.

The number of terms in the expansion of $(x+a{)}^n+(x-a{)}^n$ are:

• $(n+2)/2$ if $n$ is even
• $(n+1)/2$ if $n$ is odd.

1. Computing the coefficients of a binomial expansion
2. Calculating the probability of an event occurring in a Bernoulli trial

The Binomial theorem is used in Mathematics to determine the remainder and the digits of a number.