Binomial Theorem - Formula, Expansion and Problems

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 Binomial Theorem. Learn about all the details about binomial theorem like its definition, properties, applications, etc

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
nΣr=0 nCr xn – r · y+ nCr xn – r · y+ …………. +  nCn-1 x · yn – 1 + nC· yn
i.e.(x + y)n = nΣr=0 nCr xn – r · yr where,
Binomial Theorem
Illustration 1: Expand (x/3 + 2/y)4
Sol:
Binomial Theorem Example Question 1
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(x+ 10 x3 y+ 5xy4)
Now (√2 + 1)+ (√2 − 1)= 2[(√2)+ 10(√2)3(1)+ 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, … .., nCn are called binomial coefficients and also represented by C0, C1, C2, ….., Cn
  • The binomial coefficients which are equidistant from the beginning and from the ending are equal i.e. nC= nCn, nC= nCn-1 , nC= nCn-2 ,….. etc.
To find binomial coefficients we can use Pascal’s Triangle also.
Pascals Triangle
Finding binomial coefficients with Pascal’s Triangle
Some other useful expansions:
  • (x + y)+ (x−y)= 2[C0 x+ C2 xn-1 y+ C4 xn-4 y+ …]
  • (x + y)– (x−y)= 2[C1 xn-1 y + C3 xn-3 y+ C5 xn-5 y+ …]
  • (1 + x)n  nΣr-0 nC. x= [C+ C1 x + C2 x+ … Cn xn]
  • (1+x)+ (1 − x)= 2[C0 + C2 x2+C4 x+ …]
  • (1+x)− (1−x)= 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.

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+ C+ C+ … + C= 2n
  • C+ C+ C+ … = C+ C+ C+ … = 2n-1
  • C– C+ C– C+ … +(−1)n . nC= 0
  • nC+ 2.nC+ 3.nC+ … + n.nC= n.2n-1
  • C− 2C+ 3C− 4C+ … +(−1)n-1 C= 0 for n > 1
  • C0+ C1+ C2+ …Cn= [(2n)!/ (n!)2]
Illustration: If (1 + x)15 = a0 + a1x + . . . . .  + a15 x15 then, find the value of 
Sol:
Binomial Theorem Illustration
= C1/C0 + 2 C2/C1+ 3C3/C+ . . . . + 15 C15/C14
= 15 + 14 + 13 + . . . . . + 1 = [15(15+1)]/2 = 120

erms in the Binomial Expansion

In binomial expansion, it is often asked to find the middle term or th general term. Here, 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)= nC0 x+ nC1 xn-1 . y + nC2 xn-2 . y+ … + nCn yn
General Term = Tr+1 = nCr xn-r . yr
  • General Term in (1 + x)n is nCxr
  • In the binomial expansion of (x + y)n , the rth 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 =T10 – 4 + 2 = T8 = 10C(2x)(−1/x2)= −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– x3)2n
Sol:
(1 − 3x + 3x– x3)2n = [(1 − x)3]2n = (1 − x)6n
Middle Term = [(6n/2) + 1] term = 6nC3n (−x)3n
Determining a Particular Term:
  • In the expansion of(axp + b/xq)the coefficient of xm is the coefficient of Tr+1 where r = [(np−m)/(p+q)]
  • In the expansion of (x + a)n, Tr+1/Tr = (n – r + 1)/r . a/x

Independent Term

The term Independent of in the expansion of [ax+ (b/xq)]n is
Tr+1 = nCr an-r br, 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= 20
Illustration: Find the independent term in the expansion of:
Example Questions from Binomial Theorem
Sol:
(x1/3 + 1 – 1 – 1/√x)10 = (x1/3 – 1/√x)10
r = [10(1/3)]/[1/3+1/2] = 4
 T10C4 = 210

Numerically greatest term in the expansion of (1+x)n:

  • If [(n+1)|x|]/[|x|+1] = P, is a positive integer then Pth term and (P+1)th terms are numerically 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 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.
T6 + 1 = 10C. (−3x)= 10C6 . (3/2)6

Ratio of Consecutive Terms/Coefficients:

Coefficients of xr and xr + 1 are nCr – 1 and nCr respectively.
(nCr / nCr – 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 + 2)th and (r + 3)th be the three consecutive terms.
Then nC: nCr + 1 : nCr + 2 = 1:7:42
Now (nCr / nCr – 1) = (1/7)
(nCr / nCr – 1) = (1/7) ⇒ [(r + 1)/(n − r)] = (1/7) ⇒ n−8r = 7 → (1)
And,
(nCr / nCr – 1) = (7/42) ⇒ [(r+2)/(n – r – 1)] =(1/6) ⇒ n−7r = 13 → (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) = [2(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)= (170 − 1)5
= 5C0 (170)− 5C(170)+ 5C(170)− 5C(170)+ 5C(170) − 5C5
= 5C(170)− 5C(170)+ 5C(170)− 5C(170)+ 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 11+ 911 is divisible by 10.
Sol:
11+ 911 = (10 + 1)+ (10 − 1)11
= (9C. 10+ 9C. 10+ … 9C9) + (11C. 1011 − 11C. 1010 + … −11C11)
= 9C. 10+ 9C. 10+ … + 9C. 10 + 1 + 1011 − 11C. 1010 + … + 11C10 . 10−1
= 10[9C. 10+ 9C. 10+ … + 9C+ 11C. 1010 − 11C. 10+ … + 11C10]
= 10K, which is divisible by 10.
Formulae:
  • The number of terms in the expansion of (x+ x+ … xr)n is (n + r − 1)Cr – 1
  • Sum of the coefficients of (ax + by)n is (a + b)n
If f(x) = (a+ a1x + a2x+ …. + amxm)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
Binomial Theorem for any Index
Proof:
Let f(x) = (1 + x)n = a+ a1 x + a2 x+ … +ar x+ … (1)
f(0) = (1 + 0)= 1
Differentiating (1) w.r.t. x on both sides, we get
n(1 + x)n – 1
= a1 + 2a2 x + 3a3 x+ 4a4 x+ … + rar xr – 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
= 2a+ 6a3 x + 12a4 x+ … + r(r−1) ar xr – 2 + … (3)
Put x = 0, we get a= [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) ar xr – 3 + …
Put x = 0, we get a= [n(n−1)(n−2)] / 3!
Similarly, we get a= [n(n−1)(n−2)(n−3)] / r! and so on
∴ a= [n(n−1)(n−2)…(n−r+1)] / r!
Putting the values of a0, a1, a2, a3, …, ar obtained in (1), we get
(1 + x)= 1 + nx + [{n(n−1)} / 2!] x2 + [{n(n − 1)(n − 2)} / 2!] x+ … + [{n(n − 1)(n − 2) … (n – r + 1)}/ r!] x+ …

Binomial theorem for Rational Index

The number of rational terms in the expression of (a1/l + b1/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:
Tr + 1 = 100Cr (8√5)100 – r . (6√2)= 100C. 5[(100 – r)/8] .2r/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+ x+ … + x+ … ∞
  • (1 + x)-1 = 1 – x + x– x+ … (−1)xr+ … ∞
  • (1 − x)-2 = 1 + 2x + 3x2 − 4x3 + … + (r + 1)xr + … ∞
  •  (1 + x)-2 = 1 − 2x + 3x− 4x+ … + (−1)(r + 1)x+ … ∞
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
=nr = 0nCr xn – r ar, n∈N
nr = 0 [n! / (n − r)!r!] xn – rar
nr + s =n [n! / r!s!] xs ar, where s = n – r.
This result can be generalized in the following form:
(x1 + x2 + … +xk)n
= ∑r1 + r2 + …. + rk = n [n! / r1!r2!…rk!] x1r1 x2r2 …xkrk
The general term in the above expansion is
[(n!) / (r1! r2! r3! … rk!)] x1r1 x2r2 x3r3… xkrk
The number of terms in the above expansion is equal to the number of non-negative integral solution of the equation.
r1+r+ … + r= n, because each solution of this equation gives a term in the above expansion. The number of such solutions is n + k – 1Ck −1.
PARTICULAR CASES
Case-1:
 Multinomial Theorem Case 1
The above expansion has n+3-1C3-1 = n + 2C2 terms.
Case-2: 
Multinomial Theorem Case 2
There are n + 4 – 1C4 – 1 = n + 3C3 terms in the above expansion.
REMARK: The greatest coefficient in the expansion of (x+ x+ … + xm)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!) xP1 yP2 zP3
Where P1 + P2 + P3 = 10 and 0 ≤ P1, P2, P3 ≥ 10
In general,
(x+ x+ … xr)= ∑ (n!) / (P1! P2! … Pr!) xP1 xP2 … xPr
Where P1 + P2 + P3 + … + Pr = n and 0 ≤ P1, P2, … Pr ≥ n
Number of Terms in the Expansion of (x1 + x2 + … + xr)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, x…., xn such that the sum of powers is always “n”.
Number of non-negative integral solutions of x+ x+ … + x= n is n +r – 1Cr – 1.
For example, number of terms in the expansion of (x + y + z)3 is 3 + 3 -1C3 – 1 =  5C= 10
As in the expansion, we have terms such as
As xy0 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 – 1C3 – 1 n + 2C2.
Number of terms in (x + y + z + w)n is n + 4 – 1C4 – 1 = n + 3C3 and so on.

Problems on Binomial Theorem

Question 1: If the third term in the binomial expansion of   equals 2560, find x.
Solution:
⇒ (log2x)2 = 4
⇒ log2x = 2 or -2
⇒ x = 4 or 1/4.
Question 2: Find the positive value of λ for which the coefficient of xin the expression x2[x + (λ/x2)]10 is 720.
Solution:
⇒ x[10Cr . (√x)10-r . (λ/x2)r] = x2 [10C. λr . x(10-r)/2 . x-2r]
= x2 [10C. λr . 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., xx1+8 x2+7 x3+6 x4+5, x1+3+5, x2+3+4
∴ The coefficient of x= 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 Tr-1, Tr, Tr+1 are three consecutive terms of (1 + x)n+5
⇒  Tr-1 = (n+5) Cr-2 . xr-2
⇒  Tr = (n+5) Cr-1 . xr-1
⇒  Tr+1 = (n+5) Cr . xr
Given
(n+5) Cr-2 : (n+5) Cr-1 : (n+5) Cr = 5 : 10 : 14
Therefore, [(n+5) Cr-2]/5= [(n+5) Cr-1]/10 = (n+5) Cr/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 x100100C 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 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.
Solution:
Binomial Theorem Question
Question 11: Find the coefficient of a3b2c4in 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 + x+x3)11.
Solution:
By expanding given equation using expansion formula we can get the coefficient x4
i.e. 1 + x + x+ x= (1 + x) + x(1 + x) = (1 + x) (1 + x2)
⇒ (1 + x + x+ x3) x11 = (1+x)11 (1+x2)11
= 1+ 11Cx11Cx11Cx11Cx. . . . . . .
= 1 + 11Cx11Cx+ . . . . . .
To find term in from the product of two brackets on the right-hand-side, consider the following products terms as
= 1 × 11Cx11Cx2 × 11Cx11Cx4
11C11C× 11C11C] 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:
Tr+1 = 100Cr . 5(100 – r)/2 nr/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 27= 729
Now 2726 = (729)13 = (730 – 1)13
13C(730)13 – 13C(730)12 + 13C(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.

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