Sum of Binomial Coefficients Formula and Proof

The sum of binomial coefficients is the total of all binomial coefficients that appear in the expansion of expressions like (a + b)ⁿ for a non-negative integer n.

For example, in the expansion of (x + y)³, the binomial coefficients are 1, 3, 3, and 1. When we add these coefficients together, we get the sum of binomial coefficients: 1 + 3 + 3 + 1 = 8. This means that the sum of the binomial coefficients for n = 3 is 8.

Formula

The Sum of Binomial Coefficients Formula states that for a non-negative integer n, the sum of binomial coefficients is equal to 2ⁿ. Mathematically, it can be expressed as:

\sum_{r=0}^{n} \ ^nC_r = \ ^nC_0 + \ ^nC_1 + \ ^nC_2 + \dots + \ ^nC_{n-1} + \ ^nC_n = 2^n

Sum of Even Binomial Coefficients

The sum of even-indexed binomial coefficients for (1 + x)ⁿ is:

⅀ⁿ_{k = 0} (ⁿ _2k) = ⁿC₀ + ⁿC₂ + ⁿC₄ + ⁿC₆ + ⁿC₈ + … = 2^{n - 1}

Sum of Odd Binomial Coefficients

The sum of odd-indexed binomial coefficients for (1+x)ⁿ is:

⅀ⁿ_{k = 0} (ⁿ _{2k + 1}) = ⁿC₁ + ⁿC₃ + ⁿC₅ + ⁿC₇ + ⁿC₉ + … = 2^n-1

Note: Relationship Between Sum of Even and Odd Binomial Coefficients

• S_even + S_odd ​ = 2n
• S_even = S_odd if n is even
• S_even = S_odd + 1 if n is odd

Proof of the Sum of Binomial Coefficients

The proof for the sum of Binomial coefficients can be proved using two methods:

Method 1: Using the Principle of Induction

Base Case (n = 0):

LHS = ⁰C₀ = (0!)/(0! * 0!) = 1/1 = 1. 
RHS = 2⁰ = 1. 
Thus, LHS = RHS

For induction step: 

Let k be an integer such that k > 0 and for all r, 0 < = r < = k, where r stands to integers, the formula stands true. 
Therefore,

 ^kC₀ + ^kC₁ + ^kC₂ + ……. + ^kC_k-1 + ^kC_k = 2^k

Now, we have to prove for n = k + 1, 

^k+1C₀ + ^k+1C₁ + ^k+1C₂ + ……. + ^k+1C_k + ^k+1C_k+1 = 2^k+1

LHS = ^k+1C₀ + ^k+1C₁ + ^k+1C₂ + ……. + ^k+1C_k + ^k+1C_k+1 

(Using ⁿC₀ = 0 and ⁿ⁺¹C_r = ⁿC_r + ⁿC_r-1) 

= 1 + ^kC₀ + ^kC₁ + ^kC₁ + ^kC₂ + …… + ^kC_k-1 + ^kC_k + 1 
= ^kC₀ + ^kC₀ + ^kC₁ + ^kC₁ + …… + ^kC_k-1 + ^kC_k-1 + ^kC_k + ^kC_k 
= 2 × ? ⁿC_r 
= 2 × 2^k 
= 2^k+1 
= RHS

Thus, by induction, the formula holds for all n.

Method 2: Using Binomial Theorem Expansion

The binomial expansion is a powerful formula that expresses the expansion of the power of a binomial expression (x + y)ⁿ. It states that:

(x + y)ⁿ = ⁿC₀ xⁿ y⁰ + ⁿC₁ x^n-1 y¹ + ⁿC₂ x^n-2 y² + ……… + ⁿC_n-1 x¹ y^n-1 + ⁿC_n x⁰ yⁿ

By substituting x = 1 and y = 1, we simplify the expression:

(1 + 1)ⁿ = ⁿC₀ 1ⁿ 1⁰ + ⁿC₁ x^n-1 1¹ + ⁿC₂ 1^n-2 1² + ……… + ⁿC_n-1 1¹ 1^n-1 + ⁿC_n 1⁰ 1ⁿ

This results in:

2ⁿ = ⁿC₀ + ⁿC₁ + ⁿC₂ + ……. + ⁿC_n-1 + ⁿC_n

This equation demonstrates that the sum of all binomial coefficients for a given n equals 2ⁿ, revealing the fundamental relationship between the coefficients and the powers of two.

Solved Examples on the Sum of Binomial Coefficients

Example 1: Calculate (ⁿ ₀) + (ⁿ ₁) + (ⁿ ₂) + (ⁿ ₃) + ...(ⁿ ₁₀) for n = 10.

Solution:

Using the sum of binomial coefficient formula: (ⁿ ₀) + (ⁿ ₁) + (ⁿ ₂) + (ⁿ ₃) + ...(ⁿ _n) = 2ⁿ
Thus, 2¹⁰ = 1024

Thus, the sum is 1024.

Example 2: Find the sum of the binomial coefficients for the expansion of (x + 1)⁴.

Solution:

Given : Expression : (x + 1)⁴

Comparing the expression with ( x + a) ⁿ we get n = 4
By using the formula : Sum of binomial coefficient = 2ⁿ

The sum of the binomial coefficients is 2⁴ = 16.

Example 3: Find the sum of Odd and Even Binomial Coefficients for n = 5.

Solution:

Given: n = 5

Sum of Even Coefficient = (⁵ ₀) + (⁵ ₂) + (⁵ ₄)
⇒ Sum of Even binomial coefficient = 2^n-1 (using formula)
⇒ Sum of even coefficients: 2^5-1 = 2⁴ = 16

Sum of Odd Coefficient = (⁵₁) + (⁵₃) + (⁵₅)
⇒ Sum of odd binomial coefficient = 2^n-1 (using formula)
⇒ Sum of odd coefficients: 2^5-1 = 2⁴ = 16

Both sums are 16, confirming they are equal.

Practice Questions on Sum of Binomial Coefficients

Question 1. Evaluate \sum_{k=0}^{n} \binom{n}{k}.

Question 2. Find the value of \sum_{k=0}^{10} \binom{10}{k}.

Question 3. Evaluate \sum_{k=0}^{r} \binom{n}{k} for r = n-1.

Question 4. Compute \sum_{k=0}^{5} \binom{5}{k} \cdot k.

Question 5. Evaluate \sum_{k=1}^{n} k \cdot \binom{n}{k}.

Question 6. Find the value of \sum_{k=0}^{n} (-1)^k \binom{n}{k}.

Related Articles

• Binomial Theorem
• Corollaries of Binomial Theorem
• Pascal's Triangle
• Uses of Binomial Theorem in Daily Life