Binomial Theorem - Formula, Expansion and Problems

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 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

JEE Main 2021 Maths LIVE Paper Solutions 24-Feb Shift-1 Memory-Based

Introduction to the Binomial Theorem

The Binomial Theorem is the method of expanding an expression which 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 which contains two dissimilar terms. Ex: a + b, a³ + b³, etc.

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

(x + y)ⁿ = ⁿΣ_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[5C₀  x⁵ + 5C₂  x³ y²  +  5C₄ xy⁴]

= 2(x⁵ + 10 x³ y² + 5xy⁴)

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.
• nC₀, nC₁, nC₂, … .., nC_n are called binomial coefficients and also represented by C₀, C₁, C₂, ….., C_n
• 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.

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

Finding binomial coefficients with Pascal’s Triangle

Some other useful expansions:

• (x + y)ⁿ + (x−y)ⁿ = 2[C₀ xⁿ + C₂ x^n-1 y² + C₄ x^n-4 y⁴ + …]
• (x + y)ⁿ – (x−y)ⁿ = 2[C₁ x^n-1 y + C₃ x^n-3 y³ + C₅ x^n-5 y⁵ + …]
• (1 + x)ⁿ  = ⁿΣ_r-0 nC_r . x^r = [C₀ + C₁ x + C₂ x² + … C_n x_n]
• (1+x)ⁿ + (1 − x)ⁿ = 2[C₀ + C₂ x²+C₄ x⁴ + …]
• (1+x)ⁿ − (1−x)ⁿ = 2[C₁ x + C₃ x³ + C₅ x⁵ + …]
• 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:

• C₀ + C₁ + C₂ + … + C_n = 2ⁿ
• C₀ + C₂ + C₄ + … = C₁ + C₃ + C₅ + … = 2^n-1
• C₀ – C₁ + C₂ – C₃ + … +(−1)ⁿ . nC_n = 0
• nC₁ + 2.nC₂ + 3.nC₃ + … + n.nC_n = n.2^n-1
• C₁ − 2C₂ + 3C₃ − 4C₄ + … +(−1)^n-1 C_n = 0 for n > 1
• C₀² + C₁² + C₂² + …C_n² = [(2n)!/ (n!)²]

Illustration: If (1 + x)¹⁵ = a₀ + a₁x + . . . . .  + a₁₅ x¹⁵ then, find the value of 

Sol:

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

= 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 = nC₀ xⁿ + nC₁ x^n-1 . y + nC₂ x^n-2 . y² + … + nC_n yⁿ

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 +x²)⁵⁰

Sol:

(1 + 2x + x²)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/x²)¹⁰

Sol:

Required term =T_{10 – 4 + 2} = T₈ = 10C₇ (2x)³ (−1/x²)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 + 3x² – x³)²ⁿ

Sol:

(1 − 3x + 3x² – x³)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 = ⁿC_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 6C₃ = 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

∴ T₅ = ¹⁰C₄ = 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, T₇ is the numerically greatest term.

T_{6 + 1} = 10C₆ . (−3x)⁶ = 10C₆ . (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 application 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 7¹⁰³ is divided by 25

Sol:

(7¹⁰³ / 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 (2⁴⁰³ / 15) is (K/15), then find K.

Sol:

(2⁴⁰³ / 15) = [2³ (2⁴)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

= 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

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

∴ The last two digits are 49.

Relation Between two Numbers

Illustration: Find the larger of 99⁵⁰ + 100⁵⁰ and 101⁵⁰

Sol:

101⁵⁰ = (100 + 1)⁵⁰ = 100⁵⁰ + 50 . 100⁴⁹ + 25 . 49 . 100⁴⁸ + …

⇒ 99⁵⁰ = (100 − 1)⁵⁰ = 100⁵⁰ – 50 . 100⁴⁹ + 25 . 49 . 100⁴⁸ − ….

⇒ 101⁵⁰ – 99⁵⁰ = 2[50 . 100⁴⁹ + 25(49) (16) 100⁴⁷ + …]

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

∴ 101⁵⁰ – 99⁵⁰ > 100⁵⁰

⇒ 101⁵⁰ > 100⁵⁰ + 99⁵⁰

Divisibility Test

Illustration: Show that 11⁹ + 9¹¹ is divisible by 10.

Sol:

11⁹ + 9¹¹ = (10 + 1)⁹ + (10 − 1)¹¹

= (9C₀ . 10⁹ + 9C₁ . 10⁸ + … 9C₉) + (11C₀ . 10¹¹ − 11C₁ . 10¹⁰ + … −11C₁₁)

= 9C₀ . 10⁹ + 9C₁ . 10⁸ + … + 9C₈ . 10 + 1 + 10¹¹ − 11C₁ . 10¹⁰ + … + 11C₁₀ . 10−1

= 10[9C₀ . 10⁸ + 9C₁ . 10⁷ + … + 9C₈ + 11C₀ . 10¹⁰ − 11C₁ . 10⁹ + … + 11C₁₀]

= 10K, which is divisible by 10.

Formulae:

• The number of terms in the expansion of (x₁ + x₂ + … 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) = (a₀ + a₁x + a₂x² + …. + a_mx^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 = a₀ + a₁ x + a₂ x² + … +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}

= a₁ + 2a₂ x + 3a₃ x³ + 4a₄ x³ + … + ra_r x^{r – 1} + … (2)

Put x = 0, we get n = a₁

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

n(n − 1)(1 + x)^{n – 2}

= 2a₂ + 6a₃ x + 12a₄ x² + … + r(r−1) a_r x^{r – 2} + … (3)

Put x = 0, we get a₂ = [n(n−1)] / 2!

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

n(n − 1)(n − 2)(1 + x)^{n – 3} = 6a₃ + 24a₄ x + … + r(r − 1)(r − 2) a_r x_{r – 3} + …

Put x = 0, we get a₃ = [n(n−1)(n−2)] / 3!

Similarly, we get a₄ = [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 a₀, a₁, a₂, a₃, …, a_r obtained in (1), we get

(1 + x)ⁿ = 1 + nx + [{n(n−1)} / 2!] x² + [{n(n − 1)(n − 2)} / 2!] x³ + … + [{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 + x² + x³ + … + x^r + … ∞
• (1 + x)^-1 = 1 – x + x² – x³ + … (−1)^r x^r+ … ∞
• (1 − x)^-2 = 1 + 2x + 3x² − 4x³ + … + (r + 1)x^r + … ∞
•  (1 + x)^-2 = 1 − 2x + 3x² − 4x³ + … + (−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:

(x₁ + x₂ + … +x_k)ⁿ

= ∑_{r1 + r2 + …. + rk = n} [n! / r1!r2!…rk!] x₁^r1 x₂^r2 …x_k^rk

The general term in the above expansion is

[(n!) / (r₁! r₂! r₃! … r_k!)] x₁^r1 x₂^r2 x₃^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.

r₁+r₂ + … + 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+3-1C_3-1 = ^{n + 2}C₂ terms.

Case-2: 

There are ^{n + 4 – 1}C_{4 – 1} = ^{n + 3}C₃ terms in the above expansion.

REMARK: The greatest coefficient in the expansion of (x₁ + x₂ + … + 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 λx²y³z⁵. 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,

(x₁ + x₂ + … 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 (x₁ + x₂ + … + 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 x₁, x₂, x₃ …., x_n such that the sum of powers is always “n”.

Number of non-negative integral solutions of x₁ + x₂ + … + 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} =  ⁵C₂ = 10

As in the expansion, we have terms such as

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.

Number of terms in (x + y + z)ⁿ is ^{n + 3 – 1}C_{3 – 1} = ^{n + 2}C₂.

Number of terms in (x + y + z + w)ⁿ is ^{n + 4 – 1}C_{4 – 1} = ^{n + 3}C₃ and so on.

Binomial Theorem IIT JEE Video Lesson

Problems on Binomial Theorem

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

Solution:

⇒ (log₂^x)² = 4

⇒ log₂^x = 2 or -2

⇒ x = 4 or 1/4.

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

Solution:

⇒ x² [¹⁰C_r . (√x)^10-r . (λ/x²)^r] = x² [¹⁰C_r . λ^r . x^(10-r)/2 . x^-2r]

= x² [¹⁰C_r . λ^r . x^(10-5r)/2]

Therefore, r = 2

Hence, ¹⁰C₂ . λ² = 720

⇒ λ² = 16

⇒ λ = ±4.

Question 3: The sum of the real values of x for which the middle term in the binomial expansion of (x³/3  + 3/x)⁸ equals 5670 is?

Solution:

T₅ = ⁸C₄ × (x¹²/81) × (81/x⁴) = 5670

⇒ 70 x⁸ = 5670

⇒ x = ± √3.

Question 4: Let (x + 10)⁵⁰ + (x – 10)⁵⁰ = a₀ +a₁x + a₂ x₂ + . . . . . + a₅₀ x⁵⁰ for all x ∈R, then a₂/a₀ is equal to?

Solution:

⇒ (x + 10)⁵⁰ + (x – 10)⁵⁰:

a₂ = 2 × ⁵⁰C₂ × 10⁴⁸

a₀ = 2 × 10⁵⁰

⇒ a₂/a₀ = ⁵⁰C₂/10² = 12.25.

Question 5: Find the coefficient of x⁹ in the expansion of (1 + x) (1 + x² ) (1 + x³) . . . . . . (1 + x¹⁰⁰).

Solution:

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 x⁹ = 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! + 3¹⁸³.

Solution:

⇒ 3¹⁸³ = (3⁴)⁴⁵.3³

⇒ unit digit = 7 and 183! ends with 0

∴ Units digit of  183! + 3¹⁸³ 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[¹⁰⁰C₀ x¹⁰⁰. ¹⁰⁰C₂  x⁹⁸ . a² + . . . . . . +  ¹⁰⁰C₁₀₀ a¹⁰⁰]

∴ Total Terms = 51.

Question 9: Find the coefficient of t⁴ in the expansion of [(1-t⁶)/(1 – t)].

Solution:

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

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 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 a³b²c⁴d 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 a³b²c⁴d this implies that r₁ = 3, r₂ = 2, r₃ = 4, r₄ = 1,

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

Question 12: Find the coefficient of in the expansion of (1 + x + x² +x³)¹¹.

Solution:

By expanding given equation using expansion formula we can get the coefficient x⁴

i.e. 1 + x + x² + x³ = (1 + x) + x² (1 + x) = (1 + x) (1 + x²)

⇒ (1 + x + x² + x³) x¹¹ = (1+x)¹¹ (1+x²)¹¹

= 1+ ¹¹C₁ x² + ¹¹C₂ x² + ¹¹C₃ x³ + ¹¹C₄ x⁴ . . . . . . .

= 1 + ¹¹C₁ x² + ¹¹C₂ x⁴ + . . . . . .

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

= 1 × ¹¹C₂ x⁴ + ¹¹C₂ x² × ¹¹C₁ x² + ¹¹C₄ x⁴

= ¹¹C₂ + ¹¹C₂ × ¹¹C₁ + ¹¹C₄ ] x⁴

⇒ [55 + 605 + 330] x⁴ = 990x⁴

∴ The coefficient of x⁴ 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 = ¹⁰⁰C_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 [⁵C₀ x⁵ + ⁵C₂ x⁵ (x³ – 1) + ⁵C₄ . x . (x³ – 1)²]

Therefore, the highest power = 7.

Question 15: Find the last three digits of 27²⁶.

Solution:

By reducing 27²⁶ into the form (730 – 1)ⁿ and using simple binomial expansion we will get required digits.

We have 27² = 729

Now 27²⁶ = (729)¹³ = (730 – 1)¹³

= 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 = 1000m + 5799

Thus, the last three digits of 17²⁵⁶ are 799.

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