Binomial Theorem
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
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
$(x + y)^n = \sum_{r=0}^{n}\binom{n}{r}x^{n-r}y^r$ 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
$(x + y)^5 + (x - y)^5 = 2\binom{5}{0}x^5 + \binom{5}{2}x^3y^2 + \binom{5}{4}x^1y^4$
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 = nCn, nC1 = nCn-1, nC2 = nCn-2, … etc.
We can use Pascal’s Triangle to find binomial coefficients.
Using Pascal’s Triangle to Find Binomial Coefficients
More Useful Expansions:
$(x + y)^n + (x - y)^n = 2[C_0 x^n + C_2 x^{n-1} y^2 + C_4 x^{n-4} y^4 + \cdots]$
$(x + y)^n - (x - y)^n = 2[C_1 x^{n-1}y + C_3 x^{n-3}y^3 + C_5 x^{n-5}y^5 + \ldots]$
(1 + x)n = nΣr=0 nCr . xr = [C0 + C1 x + C2 x2 + … Cn 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 Cn = 0
nC1 + 2nC2 + 3nC3 + … + nnCn = n(2n - 1)
C₁ - 2C₂ + 3C₃ - 4C₄ + ... + (-1)ⁿ⁻¹Cⁿ = 0 for n > 1
C02 + C12 + C22 + …Cn2 = [(2n)!/ (n!)2]
Answer: If is equal to $(1+x)^{15} = a_0 + a_1x + \cdots + a_{15}x^{15}$, 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 $(x + y)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^2 + \ldots + \binom{n}{n}y^n$
General Term = ${T_r+1} = \binom{n}{r} \cdot x^{n-r} \cdot y^r$
(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 = $$\frac{6n}{2} + 1 \text{term } = 6nC3n (-x)^3n$$
Determining a Particular Term:
The coefficient of $x^m$ in the expansion of $(ax^p + b/x^q)^n$ 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 [axp + (b/xq)]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
(x1/3 + 1 - 1 - x-1/2)10 = (x1/3 - 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 ${n \choose r} - 1$ and ${n \choose r}$ 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 $$\frac{ \binom{n}{r} - 2 }{ \binom{n}{r} - 1 } = \frac{1}{7}$$
(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
- Computing powers of binomials
- Solving polynomial equations
- Computing probabilities in probability theory
- 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 $(x_1 + x_2 + \dots + x_r)^n$ is ${n + r - 1 \choose r - 1}$.
The sum of the coefficients of $(ax + by)^n$ is $(a + b)^n$
If $$f(x) = (a_0 + a_1x + a_2x^2 + \ldots + a_mx^m)^n$$ 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 $f(x) = (1 + x)^n = a_0 + a_1 x + a_2 x^2 + \dots + a_r x^r + \dots$ (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
$n(n - 1)(1 + x)^n - 2$
$2a^2 + 6a^3x + 12a^4x^2 + \ldots + r(r-1)arx^r - 2 + \ldots (3)$
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)arxr - 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
$(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{2!}x^3 + \dots + \frac{n(n-1)(n-2)\dots(n-r+1)}{r!}x^r + \dots$
The Binomial Theorem for a Rational Index
The number of rational terms in the expression of $$\left(\frac{a_1}{l} + \frac{b_1}{k}\right)^n$$ 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
$\frac{1}{1 - x} = 1 + x + x^2 + x^3 + … + x^r + … \infty$
$(1+x)^{-1} = 1 - x + x^2 - x^3 + \cdots (-1)^rx^r + \cdots \infty$
$(1 - x)^{-2} = 1 + 2x + 3x^2 - 4x^3 + \dots + (r + 1)x^r + \dots \infty$
$(1 + x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + \ldots + (-1)^r(r+1)x^r + \ldots \infty$
![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!)] xsr, where s = n – r.
This result can be generalized as follows:
(x1 + x2 + … + xk)<sup>n</sup>
$$\sum_{i=1}^{k}r_i = n \left[ \frac{n!}{r_1!r_2!\cdots r_k!}x_1^{r_1}x_2^{r_2} \cdots x_k^{r_k} \right]$$
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 $$r_1 + r_2 + \dots + r_k = n$$ is $$\binom{n+k-1}{k-1}$$, because each solution of this equation gives a term in the above expansion.
Particular Cases
Case-1:
The above expansion has $$n+3-1\binom{3-1}{2} = n + 2\binom{2}{2}$$ 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 $\frac{n!}{q!m - r\left(q+1\right)!^r}$, 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 $(x_1 + x_2 + \dots + x_r)^n$
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 $$x_1 + x_2 + \dots + x_r = n$$ is $$\binom{n+r-1}{r-1}$$
The number of terms in the expansion of $(x + y + z)^3$ is $\binom{3}{3} - \binom{3}{1} + 1 = \binom{5}{2} = 10$
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 $\binom{n+3-1}{3-1} - 1 = \binom{n+2}{2}$.
The number of terms in $(x + y + z + w)^n$ is $n + 4 - \binom{4}{1} - 1 = n + 3\binom{3}{1}$ 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, $$\binom{10}{2}\lambda^2 = 720$$
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 $(x + 10)^{50} + (x - 10)^{50} = a_0 + a_1x + a_2x^2 + \dots + a_{50}x^{50}$ 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:
This is a header
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 xr in (1 – x)-n = (r + n – 1) Cr
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:
This is a heading
Solution:
This is a heading
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:
This is a heading
Solution:
This is a heading
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 $r = 0, 1, 2, \ldots, 100$
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: $(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k}$
$(x + y)^n = \sum_{r=0}^{n} \binom{n}{r} x^{n-r}y^r$
What is the General Term in a Binomial Expansion?
The general term of a binomial expansion is Tr+1 = nCr xn-r yr.
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.
- Computing the coefficients of a binomial expansion
- 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.
JEE Study Material (Mathematics)
- 3D Geometry
- Adjoint And Inverse Of A Matrix
- Angle Measurement
- Applications Of Derivatives
- Binomial Theorem
- Circles
- Complex Numbers
- Definite And Indefinite Integration
- Determinants
- Differential Equations
- Differentiation
- Differentiation And Integration Of Determinants
- Ellipse
- Functions And Its Types
- Hyperbola
- Integration
- Inverse Trigonometric Functions
- Limits Continuity And Differentiability
- Logarithm
- Matrices
- Matrix Operations
- Minors And Cofactors
- Properties Of Determinants
- Rank Of A Matrix
- Solving Linear Equations Using Matrix
- Standard Determinants
- Straight Lines
- System Of Linear Equations Using Determinants
- Trigonometry
- Types Of Matrices