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241 LearnersLast updated on October 21, 2025

The binomial theorem is a formula used for expanding terms. It includes the sum of terms like ‘a’ and ‘b’ with powers without directly multiplying the power. It provides formulas for expressions to calculate easily. Here, in this article, we will be discussing the binomial theorem.
The binomial theorem was initially discussed in the fourth century BC by the significant Greek mathematician Euclid. The binomial theorem offers the means to expand the algebraic equation (x + y)n into a sum of terms involving individual exponents of variables x and y.
When you expand a binomial, each part of the expanded form has a number in front of it – that number is the coefficient. It's just a multiplier for each term in the expansion.
 
The binomial theorem allows one to expand any non-negative power of a binomial (x + y) into a sum of the form.
(x + y)n = nC0 xny0 + nC1 xn-1y1 + nC2 xn-2 y2 + ... + nCn-1 x1yn-1 + nCn x0yn
n ≥ 0 is an integer; each nCk is a positive integer, sometimes referred to as a binomial coefficient.
 
The binomial theorem provides a quick formula to expand expressions without manual multiplication. When you use this theorem, you get a specific pattern for the expanded form of (x + y)n:
 
(x + y)n = nC0 xny0 + nC1 xn-1y1 + nC2 xn-2 y2 + ... + nCn-1 x1yn-1 + nCn x0yn
 
In simpler terms, the binomial theorem's formula is the key to taking an expression and writing it out as a sum of individual terms.
For any positive whole number n, the expansion of (a + b)n = ∑nr = 0nCr an-rbr, where, a and b are real numbers, and 0 < r ≤ n.
This formula extends the binomial expressions, including (x + a)10, (2x + 5)3, (x - (1/x))4, and so forth. The binomial theorem formula helps a binomial raised to a given degree to expand easily. Let us grasp the formula for the binomial theorem and its application in the next parts.
The Binomial Theorem states that for any real or complex numbers x and y, and for any non-negative whole number n (belonging to the set n ∈ 𝑁 = {0, 1, 2, ...}), a specific expansion holds true,
⇒ (x + y)n = ∑nk = 0nk xn-kyk
where the binomial coefficient nk is indeed calculated as:
nk = n!/k! (n - k) !
The representation is accurate and clearly shows the expansion of (x + y)n as a sum of terms involving powers of x and y, along with the binomial coefficients.
 
Let x, y, and n be from N. Let's use mathematical induction to demonstrate the binomial theorem formula. For n = 1, n = 2, for n = k ≥ 2, and for n = k + 1, it is sufficient to show.
It is obvious that (x + y)1 = x + y and
(x + y)2 = (x + y) (x + y)
= x2 + xy + xy + y2 (using distributive property)
= x2 + 2xy + y2
For n = 1 and n = 2, the outcome is hence true.
Suppose k to be a positive integer. Let us show that for k ≥ 2 the outcome is correct.
Assuming (x + y)n = ∑nr=0nCr xn-ryr
(x + y)k = ∑krkCr xk-ryr
⇒ (x+y)k = kC0 xky0 + kC1 xk-1y1 + kC2 xk-2 y2 + ... + kCr xk-ryr +....+ kCk x0yk
⇒ (x+y)k = xk + kC1 xk-1y1 + kC2 xk-2 y2 + ... + kCr xk-ryr +....+ yk
Thus, the result is true for n = k ≥ 2.
Think now about the expansion for n = k + 1.
(x + y) k+1 = (x + y) (x + y)k
= (x + y) (xk + kC1 xk-1y1 + kC2 xk-2 y2 + ... + kCr xk-ryr +....+ yk)
= xk+1 + (1 + kC1)xky + (kC1 + kC2) xk-1y2 + ... + (kCr-1 + kCr) xk-r+1yr + ... + (kCk-1 + 1) xyk + yk+1
= xk+1 + k+1C1xky + k+1C2 xk-1y2 + ... + k+1Cr xk-r+1yr + ... + k+1Ck xyk + yk+1
[Because nCr + nCr-1 = n+1Cr]
For n = k+1, the outcome is hence true. Mathematical induction proves this true for all positive numbers "n." Thus proved.
 
In the expanded form of (x + y)ⁿ, the numerical parts multiplying the powers of x and y are called binomial coefficients.
The binomial coefficients show themselves as nC0, nC1, nC2..... One can derive the binomial coefficients by means of the Pascal’s triangle or the combinations' formula.
Pascal's triangle displays the binomial coefficients that appear in binomial expansions. The binomial theorem formula sums together this evolved pattern.
 
Pascal's triangle visually reveals a pattern in the values of binomial coefficients. It's a triangular arrangement of these coefficients, named after Blaise Pascal. The edges of this triangle are all 1s, and each internal number is the sum of the two numbers directly above it.
The binomial theorem provides a formula based on combinations to directly calculate the value of these binomial coefficients that appear in the expansions of binomials. Under this situation, the numerous ways to select r variables from the given n variables form the combinations. The formula to determine the combinations of r things selected from n distinct objects is nCr = n! / [r! (n - r)!].
Here, the coefficients possess these specific characteristics.
 
nCn = nC0 = 1
nC1 = nCn-1 = n
nCr = nCr-1
 
By substituting x = 1 and y = 1 into the binomial expansion of (x + y)ⁿ, we can easily derive several properties of binomial coefficients. These characteristics reveal interesting relationships between the coefficients in the expansion.
 
C1 + C2 + C3 + C4 + .......Cn = 2n
C0 + C2 + C4+ .... = C1 + C3 + C5 + ....... = 2n-1
C0 - C1+ C2 - C3 + C4 - C5 + .... = 0
C1 + 2C2 + 3C3 + 4C4 + .......nCn = n2n-1
C1 - 2C2 + 3C3 - 4C4 + .......(-1)nnCn = 0
C12 + C22 + C32 + C42 + .......Cn2 = (2n)! / (n!)2
 
Here are some effective tips and tricks to help students learn and apply the Binomial Theorem confidently:
 
Although the Binomial Theorem is an effective algebraic tool, it can be simply incorrect if important concepts are ignored. The following are the most usual mistakes made by students, along with tips on how to prevent them:
In real life, binomial theorems are used in various fields like finance, statistics, project management, genetics. Here are a few examples of real life applications of Binomial Theorem.
 
	 
Expand x+23
x3+6x2+12x+8.
Step 1: Identify the values a=x, b=2, and n=3.
(x+y)n = ∑nk=0nCk xn-kyk = ∑nk=0nCk xkyn-k
 
Step 2: Based on the above binomial formula, the expansion will be,
x+23=k=033kx3-k2k
 
Step 3: Now, each step will be calculated:
For k=0
30x3-020=1x31=x3
For k=1:
31x3-121=3x22=6x2
For k=2:
2x3-222=3x14=12x
For k=3:
33x3-323=1x08=8
 
Step 4: Lastly, bring all the terms together:
x+23=x3+6x2+12x+8
 
Therefore, the final answer will be x3+6x2+12x+8.
Determine the coefficient of x4 from the expression 2x-36.
2160
The general term for it a+bn is:
 Tk + 1=nkan-kbk  
             
Step 1: Here, a=2x, b= -3, and n=6 using the formula, the general term will be;  
                             
Tk-1=6k(2x)6-k(-3)k=6k26-kx6-k-3k
Step 2: Set the exponent of 𝑥 to 4 since we need the term where the power of 𝑥 is 4:
 6-k=4
k=6-4=2
Step 3: Substitute the value k=2, and we get
T2+1=6226-2x6-2-32=6224x49
Step 4: Calculate the term:
62=6!2! 4!=15,
24=16,
-32=9
So, T3=15169x4=2160 x 4
Therefore, the coefficient of x 4 will be 2160.      
Expand 1+x5
1 + 5x + 10 x 2 + 10 x 3 + 5 x 4 + x5
 Step 1: Let us take a=1, b=x, and n=5 the expansion will be
1+x5= k=055k15-kxk
=k=055k xk
Step 2: Calculate each term:
For k=0
50x0=1
For k=1
51x1=5x    
 
For k=2
52x2=10x2  
  
For k=3
 
53x3=10x3
For k=4
54x4=5x4
For k=5
 
55x5=x5
 
Step 3: By combining all the terms, we get
1+x5=1+5x+10x2+10x3+5x4+x5
Therefore, 1+5x+10x2+10x3+5x4+x5 is the final answer.
Find the middle term in x-2y6
-160 x 3y3
If we consider n=6 (which is an even number), there will be two middle terms n2=3 and n2+1=4.
Using the general term:
Tk+1=6kx6-k-2yk
=6kx6-k(-2)kyk
So, if we solve for k=3, we get,
T3+1=63x6-3(-2)3y3
=20x3-8y3
=-160x3y3
Therefore, the middle term will be -160x3y3.
Determine the 4th term in the expansion of 3x+25
720x2
The general term is Tr+1=5r3x5-r(2)r
Let’s set r=3, then the 4th term will be:
T3+1=533x5-3(2)3
T4=533x5-3(2)3
=10(3x)28
=720x2
Therefore, the 4th term will be 720x2.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.






