In (x + y )ⁿ or ( 1 + x )ⁿ, we know the binomial coefficients are :

ⁿC₀,ⁿC₁, ⁿC₂, ⁿC₃,….. ⁿC_r… ⁿC_n.

In short, the binomial coefficients are also written as

C₀, C₁, C₂, C₃,….. C_r… C_n.

Let us now study a few salient properties of binomial coefficients:

Consider

( 1 + x )ⁿ = ⁿc₀ + ⁿc₁ . x + ⁿc₂ .x² + ⁿc₃ . x³ +……..+ⁿcn . xⁿ………………(E)

To find the sum of the binomial coefficients ⁿC₀,ⁿC₁, ⁿC₂, ⁿC₃,….. ⁿC_r… ⁿC_n in (E) above,

Put x = 1.

1. 2ⁿ = ⁿc₀ +ⁿc₁ + ⁿc₂ + …………. + ⁿc_n

2. Coefficients are odd or even based on the value of r.

If the r values of terms are odd, then binomial coefficients of such terms are called odd coefficients and if the r values of terms are even then binomial coefficients of such terms are called even coefficient. Therefore

C₁, C₃, C₅ and so on are odd coefficients and C₂, C₄ , C₆ and so on are even coefficients

In (1 + x )ⁿ =ⁿc₀ + ⁿc₀ . x + ⁿc₂ . x² + ……… + ⁿc_n . xⁿ

let us substitute -1 in x
( 1 – 1 )ⁿ = ⁿc₀ + ⁿc₁ . ( -1 ) + ⁿc₂ . (- 1 )² + ⁿc₃ .(-1)³ + ⁿc₄ . (-1)⁴ + ……+ ⁿc_n .
(-1)ⁿ

0 = ⁿc₀ - ⁿc₁ + ⁿc₂ – ⁿc₃ + ⁿc₄ +…………..+ ⁿc_n (-1)ⁿ

Thus, ⁿc₀ + ⁿc₂ + ⁿc₄ +…………… = ⁿc₁ + ⁿc₃ + ⁿc₅ + ………….

2. c₀ + c₂ + c₄ +……….. = c₁ + c₃ + c₅ + ………….

Because the sum of the both the odd and even binomial coefficients is equal to 2ⁿ, so the sum of the odd coefficients = ½ (2ⁿ ) = 2^{n – 1} , and

Sum of the even binomial coefficients = ½ (2ⁿ) = 2^{n – 1}

Thus, sum of the even coefficients is equal to the sum of odd coefficients.

In this way, we can derive several more properties of binomial coefficients by substituting suitable values for x and others in the binomial expansion.

12. Number of terms in the following expansions: