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Before learning binomial expansion formulas, let us recall what is a “binomial”. A binomial is an algebraic expression with two terms. For example, a + b, x – y, etc are binomials. We have a set of algebraic identities to find the expansion when a binomial is raised to exponents 2 and 3. For example, (a + b)2 = a2 + 2ab + b2. But what if the exponents are bigger numbers? It is tedious to find the expansion manually. The binomial expansion formula eases this process. Let us learn the binomial expansion formula along with a few solved examples.
What Are Binomial Expansion Formulas?
As we discussed in the earlier section, the binomial expansion formulas are used to find the powers of the binomials which cannot be expanded using the algebraic identities. The binomial expansion formula involves binomial coefficients which are of the form

(or) nCk and it is calculated using the formula,

The binomial expansion formula is also known as the binomial theorem. Here are the binomial expansion formulas.

Binomial Expansion Formula of Natural Powers
This binomial expansion formula gives the expansion of (x + y)n where ‘n’ is a natural number. The expansion of (x + y)n has (n + 1) terms. This formula says:


Here we use nCk formula to calculate the binomial coefficients which says nCk = n! / [(n – k)! k!]. By applying this formula, the above binomial expansion formula can also be written as,


Note: If we observe just the coefficients, they are symmetric about the middle term. i.e., the first coefficient is the same as the last one, the second coefficient is as same as the one that is second from the last, etc.
Binomial Expansion Formula of Rational Powers
This binomial expansion formula gives the expansion of (1 + x)n where ‘n’ is a rational number. This expansion has an infinite number of terms.
(1 + x)n = 1 + n x + [n(n – 1)/2!] x2 + [n(n – 1)(n – 2)/3!] x3 +…
Note: To apply this formula, the value of |x| should be less than 1.
Binomial Theorem Statement
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 applications in Algebra, probability, etc.
Binomial Expression: A binomial expression is an algebraic expression that contains two dissimilar terms. Eg.., a + b, a3 + b3, etc.
Some other useful expansions:
- (x + y)n + (x−y)n = 2[C0 xn + C2 xn-1 y2 + C4 xn-4 y4 + …]
- (x + y)n – (x−y)n = 2[C1 xn-1 y + C3 xn-3 y3 + C5 xn-5 y5 + …]
- (1 + x)n = nΣr-0 nCr . xr = [C0 + C1 x + C2 x2 + … Cn xn]
- (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 is (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 is (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:
- C0 + C1 + C2 + … + Cn = 2n
- C0 + C2 + C4 + … = C1 + C3 + C5 + … = 2n-1
- C0 – C1 + C2 – C3 + … +(−1)n . nCn = 0
- nC1 + 2.nC2 + 3.nC3 + … + n.nCn = n.2n-1
- C1 − 2C2 + 3C3 − 4C4 + … +(−1)n-1 Cn = 0 for n > 1
- C02 + C12 + C22 + …Cn2 = [(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
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 xn-1 . y + nC2 xn-2 . y2 + … + nCn yn
General Term = Tr+1 = nCr xn-r . yr
- General Term in (1 + x)n is nCr xr
- In the binomial expansion of (x + y)n, the rth term from the end is (n – r + 2)th.
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 = 6nC3n (−x)3n
Determining a Particular Term:
- In the expansion of (axp + b/xq)n 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
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 = 10C7 (2x)3 (−1/x2)7 = −960x-11
Independent Term
The term Independent of in the expansion of [axp + (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 6C3 = 20
Illustration: Find the independent term in the expansion of:

Sol:

(x1/3 + 1 – 1 – 1/√x)10 = (x1/3 – 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 the Pth 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.
T6 + 1 = 10C6 . (−3x)6 = 10C6 . (3/2)6
Ratio of Consecutive Terms/Coefficients:
The 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)th and (r + 1)th be the three consecutive terms.
Then, the given ratio is 1:7:42
Now (nCr-2 / nCr – 1) = (1/7)
(nCr-2 / nCr – 1) = (1/7) ⇒ [(r – 1)/(n − r+2)] = (1/7) ⇒ n−8r+9=0 → (1)
And,
(nCr-1 / nCr) = (7/42) ⇒ [(r)/(n – r +1)] =(1/6) ⇒ n−7r +1=0 → (2)
From (1) & (2), n = 55
Applications of Binomial Theorem
The binomial theorem has a wide range of applications in Mathematics, like finding the remainder, finding the digits of a number, etc. The most common binomial theorem applications are as follows:
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 + … xr)n is (n + r − 1)Cr – 1
- The sum of the coefficients of (ax + by)n is (a + b)n
If f(x) = (a0 + a1x + a2x2 + …. + 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,

Proof:
Let f(x) = (1 + x)n = a0 + a1 x + a2 x2 + … +ar xr + … (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 + … + 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
= 2a2 + 6a3 x + 12a4 x2 + … + r(r−1) ar xr – 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) ar xr – 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
∴ ar = [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)n = 1 + nx + [{n(n−1)} / 2!] x2 + [{n(n − 1)(n − 2)} / 2!] x3 + … + [{n(n − 1)(n − 2) … (n – r + 1)}/ r!] xr + …
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)r = 100Cr . 5[(100 – r)/8] .2r/6.
∴ r = 12,36,60,84
The number of rational terms = 4
The number of irrational terms = 101 – 4 = 97
Binomial Theorem for Negative Index
1. If the rational number and -1 < x <1, then,
- (1 − x)-1 = 1 + x + x2 + x3 + … + xr + … ∞
- (1 + x)-1 = 1 – x + x2 – x3 + … (−1)r xr+ … ∞
- (1 − x)-2 = 1 + 2x + 3x2 − 4x3 + … + (r + 1)xr + … ∞
- (1 + x)-2 = 1 − 2x + 3x2 − 4x3 + … + (−1)r (r + 1)xr + … ∞

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 the binomial theorem, we have
(x + a)n
=n∑r = 0nCr xn – r ar, n∈N
= n∑r = 0 [n! / (n − r)!r!] xn – rar
= n∑r + s =n [n! / r!s!] xs ar, where s = n – r.
This result can be generalised 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 solutions of the equation.
r1+r2 + … + rk = 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:

The above expansion has n+3-1C3-1 = n + 2C2 terms.
Case-2:

There are n + 4 – 1C4 – 1 = n + 3C3 terms in the above expansion.
REMARK: The greatest coefficient in the expansion of (x1 + x2 + … + 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,
(x1 + x2 + … xr)n = ∑ (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, x3 …., xn, such that the sum of powers is always “n”.
The number of non-negative integral solutions of x1 + x2 + … + xr = n is n +r – 1Cr – 1.
For example, the number of terms in the expansion of (x + y + z)3 is 3 + 3 -1C3 – 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.
The number of terms in (x + y + z)n is n + 3 – 1C3 – 1 = n + 2C2.
The 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 x2 in the expression x2[√x + (λ/x2)]10 is 720.
Solution:
⇒ x2 [10Cr . (√x)10-r . (λ/x2)r] = x2 [10Cr . λr . x(10-r)/2 . x-2r]
= x2 [10Cr . 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., 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 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 equations (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
∴ The 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 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:

Question 11: Find the coefficient of a3b2c4d in the expansion of (a-b-c+d)10.
Solution:
Expand (a – b – c + d)10 using the multinomial theorem, and by using the coefficient property, we can obtain the required result.
Using the 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 the given equation using the 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 the 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:
Tr+1 = 100Cr . 5(100 – r)/2 nr/4
Where r = 0, 1, 2, . . . . . . , 100
r must be 0, 4, 8, … 100
The 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 the 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.
Examples Using Binomial Expansion Formulas
Example 1: Find the expansion of (a + b)3.
Solution:
To find: (a + b)3
Using binomial expansion formula,
(x + y)n = nC0 xn y0 + nC11 xn – 1 y1 + nC2 xn-2 y2 + nC3 xn – 3 y3 + … + nCn−1 x yn – 1 + nCn x0yn

= ( 3! / [(3-0)!0!] ) a3 + ( 3! / [(3-1)!1!] ) a(3 – 1) b + ( 3! / [(3-2)!2!] ) a(3 – 2) b2 + ( 3! / [(3-3)!3!] ) a(3 – 3) b3

Example 2: Find the expansion of (x + y)6.
Solution:
Using the binomial expansion formula,

= ( 6! / [(6-0)!0!] ) x6 + ( 6! / [(6-1)!1!] ) x5 y + ( 6! / [(6-2)!2!] ) x4y2 + ( 6! / [(6-3)!3!] ) x3y3 + ( 6! / [(6-4)!4!] ) x2y4 + ( 6! / [(6-5)!5!] ) xy5 + ( 6! / [(6-6)!6!] ) y6
= x6 + 6x5 y + 15x4 y2 + 20x3 y3 + 15x2 y4 + 6x y5 + y6
Answer: (x + y)6 = x6 + 6x5 y + 15x4 y2 + 20x3 y3 + 15x2 y4 + 6x y5 + y6.
Example 3: Find the expansion of (3x + y)1/2 upto the first three terms using the binomial expansion formula of rational exponents where

Solution:
(3x + y)1/2 = 3x (1 + y/(3x))1/2
Comparing (1 + y/(3x))1/2 with (1 + x)n, we have x =y/(3x) and n = 1/2.
The expansion of (1 + y/(3x))1/2 upto the first three terms using the binomial expansion formula is,
1 + n x + [n(n – 1)/2!] x2 = 1 + (1/2) (y / (3x)) + [(1/2) ((1/2) – 1)/2!] (y / (3x))2
= 1 + y / (6x) – y2 / (72x2)
Thus, the expansion of 3x (1 + y/(3x))1/2 upto the first three terms is:
3x [ 1 + y / (6x) – y2 / (72x2) ] = 3x + y / 2 – y2 / (24x)
Answer: (3x + y)1/2 = 3x + y / 2 – y2 / (24x).
FAQs on Binomial Expansion Formulas
What Is Binomial Expansion In Maths?
Binomial expansion is to expand and write the terms which are equal to the natural number exponent of the sum or difference of two terms. For two terms x and y the binomial expansion to the power of n is


Here in this expansion the number of terms is equal to one more than the value of n.
What Are Binomial Expansion Formulas?
The binomial expansion formulas are used to find the expansion when a binomial is raised to a number. The binomial expansion formulas are:


where ‘n’ is a natural

- (1 + x)n = 1 + n x + [n(n – 1)/2!] x2 + [n(n – 1)(n – 2)/3!] x3 +… , when ‘n’ is a rational number and here |x| < 1.
How To Derive Binomial Expansion Formula?
The binomial expansion formula is


x0yn and it can be derived using mathematical induction. Here are the steps to do that.
- Step 1: Prove the formula for n = 1.
- Step 2: Assume that the formula is true for n = k.
- Step 3: Prove the formula for n = k.
What Are the Applications of the Binomial Expansion Formula?
The main use of the binomial expansion formula is to find the power of a binomial without actually multiplying the binominal by itself many times. This formula is used in many concepts of math such as algebra, calculus, combinatorics, etc.
How To Use the Binomial Expansion Formula?
The binomial expansion formula says the expansion of (x + y)n is


If we have to find the expansion of (3a – 2b)7, we just substitute x = 3a, y = -2b and n = 7 in the above formula and simplify.
Q1
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 = Σr=0n nCr xn – r · yr.
Q2
What is the general term in a binomial expansion?
The general term of a binomial expansion is Tr+1 = nCr xn-r yr.
Q3
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 is (n+2)/2 if n is even or (n+1)/2 if n is odd.
Q4
List two applications of the binomial theorem.
In Mathematics, the binomial theorem is used to find the remainder and also find the digits of a number.
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