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Binomial Theorem

Class 11 binomial theorem

Binomial Theorem

In mathematics, Binomial expression is that expression which contains two terms. Some popular examples would be x+y, a+b, m+n etc. We have  also often come across such binomial expressions in the forms: (x+y)², (a+b)ᶟ and so on. It is  a no-brainer for us to expand such expressions as the formulas are quite short and simple to calculate. However, when the exponent becomes higher than 5, the same task of expansion becomes quite rigorous. Hence, there came a need for a formula to work out the expansion of such higher power. The formula was derived through a theorem; hence the theorem was called “Binomial Theorem” which was first discovered by none other than the exceptional genius of Sir Isaac Newton. We derive and understand the very theorem in this chapter.

Derivation of the binomial theorem:

For any positive integer n,


We have,

(a+x)ⁿ= (a+x)×(a+x)×(a+x)……. to n factors.

Here, we notice that every time we raise the power of any term by a number, we are simply multiplying the term by itself for that much number of times. E.g. a3 = a×a×a, i.e a multiplied by itself 3 times. Since, it is a binomial operation, it is a little different. E.g. (a+x)²= (a+x)×(a+x)= a²+ a×x+a×x+x², this we know is obtained by multiplying each element of the first factor by each element of the second factor in the brackets. Now, notice that in our final answer; a²+ a×x+a×x+x², each term is simply a product of a and x, x and a, x and x or a and a where the degree of each term is always 2. With this evaluation, let us now raise our binomial expression to the power of n, to generalize it. If we have (a+x)ⁿ, each term in the expansion will simply be a product of a and x, x and a, a and a or x and x but the degree of each term will always be n. We also know that we will have (a+x) multiplied n times in the expansion. From these n factors if we take a from n factors and x from 0 factors, we get, aⁿ×x0(which we know is a term in the expansion of (a+x) where the degree is n+0=n). If we take a from (n-r) factors and x from the remaining r factors we get, 

Now any term of the expansion can be found for varying values of r. However, we still have to find out how many times our term was repeated in the expansion, like in the previous example, (a+x)²= a²+2×a×x+x², the term ax is repeated twice, to give 2×a×x as the second term. In order to solve this problem, we use the concept of our previous chapter of “Combinations”, where we allow the value of r to vary from 0 to n to get our required expansion, i.e. C(n,r) where r ranges from 0 to n.

Now, with all this in mind, let us prove the accuracy of this formula by the method of mathematical induction. 


a.   General term

The (r+1)th term of our expansion is usually called the general term because we can get any term of the expansion by providing a suitable value to r, but why (r+1)th term?


Hence,( r+1)th term gives us the term at any place where the value of r can be varied as per the requirement.

b.   Middle Term

To find the middle term we have to consider two cases.

i.                     When n is even

            In this case, we write n=2×m where m=1, 2, 3… The number of terms after expansion becomes 2×m+1 (The number of terms after expansion always increases by 1 to the power, that is in (a+x) ²= a²+2×a×x+x², in R.H.S there are 3 terms while the power is 2), which is odd. So, it has only one middle term namely, (m+1) Th term. So,

ii.                 When n is odd

We write, n=2×m-1, where m=1, 2, 3… The number of terms after expansion becomes 2×m, which is even. Now, there will be 2 middle terms mth term and (m+1)th term. So,



c.    Binomial Coefficients

What are coefficients? The numbers or constants that multiply a variable. In our theorem those numbers are computed by the method of combinations, So, C(n,0),C(n,1)……….C(n,n) are our binomial coefficients. They have the following properties:



That's all for now.