Simple Proof of Binomial Theorem for IIT-JEE

THE PROOF OF BINOMIAL THEOREM

 

Well most of the books like Arihant and all have presented the proof of Binomial theorem but either it is too difficult or the theorem is stated as such. I have given a very simple rule which makes the theorem look like a COMMON SENSE plus I had described why we have things like ⁿC_r appearing as coefficient.

Consider the binomial expansion of (a + b)n

 

So indeed we have the cases

(a + b)(a +b)………n times

 

So on multiplication we are free to form combination of  a and b so that they have total n terms. I mean to say the sum of the coefficient of a and b is n . So  the general term will be of the form

 A a^r b ^n-r    ……….1)

 

so like this many combination are possible . For example consider (a+b)2

 

so     (a +b)(a+b)

 

Let revise class 4 th concept. How do we mutliply we take one part of first bracket and mutliply it to second bracket and so on…….so we get

 

(a + b)(a + b) = aa + ab +ba + bb

 

Moreover we can say that

 “ NO TWO TERMS AFTER ADDITION WILL CONTAIN SAME COEFFICIENT OF a “ ……..2)

 

Like we can never have terms like

 

(a + b)ⁿ =  ………A a^r b ^n-r + B a^rb^n-r-1

 

So now a simple combination is need ……

 

Consider a visual picture

 

(a + b)ⁿ = (a +b) (a+ b)…… n times

 

Now friends consider each (a +b ) as a bag containing two balls marked ‘a’  and ‘b’. so we get

(a + b)ⁿ  =  { a , b } { a,b } { a,b} …… n times

 

Now all you have to do is select n balls so that “ NO 2 BALLS ARE FROM THE SAME BAG”

 

And from the result 2) if we take r balls marked as ‘a ‘ so n-r are blue so focus on one ball say ‘a’ .

 

So tell me from n balls we need to take r balls so what is the total number of ways of seleciting these balls. Yes it is the nCr so the

 

 Term a ^r b ^n-r appears nCr times so the coefficient A is nCr.

 Moreover we have terms like

      aⁿ and   bⁿ so the value of r is from 0 to n. 

 

Hence we get

 

(a + b)n = Σ_r=0 ^{n  n}C_r a^r b ^n-r