1.; ( x + y )² = x² + 2xy + y²
        2. ( x + y )³ = x³ + 3x² y + 3xy² + y³
        3. (x + y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴
        4. (x + y )⁵ = x⁵ + 5x⁴y + 10x³ y² +10x²y³ +5xy³ + y⁵

For example, in ( x + y )⁵ above, the binomial coefficients start from 1, peak up to 10, and again fall to 1.

Let us write the values of all of the six binomial coefficients in( x + y )⁵ below:

We note that the binomial coefficients always start with 1, rise to greatest value at the middle term or middle terms depending on whether index n is even or odd integer, and then again fall to 1 in the last term.

Now, let us learn a formula to find the greatest binomial coefficient:

Let the binomial expansion be in the form (1 + x ) ⁿ, where index n is a positive integer.
Two cases arise depending on whether the binomial index n is even or odd integer.

Case 1: When n is an odd integer:

Then there are two greatest binomial coefficients (these two are middle terms). They are:

ⁿc_(n+1)/2 ,ⁿc_{( n + 3 )/2}

Case 2: When n is an even integer.

Then there is only one greatest binomial coefficient (this is the only one middle term). It is: ⁿc_{( ⁿ/2 + 1)}

Eg 1. Find the greatest binomial coefficients in ( 1 + x )¹¹

Solution:
since binomial index n is 11, an odd integer, therefore the greatest binomial coefficients are two and they are:
¹¹c_{(11 + 1)/2} = ¹¹c₆
¹¹c_{(11 + 3)/2} =¹¹c₇

Eg 2: Find the greatest binomial coefficient in the binomial expansion
( 1 + x )¹⁰

Solution:
Since index n is 10; there is only one greatest binomial coefficient . And it is

¹⁰c_{(¹⁰/2) + 1} = ¹⁰c₆