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Introduction to Binomial Theorem

A Binomial Expression

Any algebraic expression consisting of only two terms is known as a binomial expression. It's expansion in power of x is shown as the binomial expansion.

For example: (i)  a + x   (ii)   a2 + 1/x2  (iii)  4x - 6y

Binomial Theorem

Such formula by which any power of a binomial expression can be expanded in the form of a series is known as binomial theorem.

It can be easily understood by examples.

        (a + x)2 = a2 + 2ax + x2

        (a + x)2 = a3 + 3a2x + 3ax2 + x3

Here, we see that the expression of (a + x)2 is simple, we just multiply (a + x) by (a + x). Expansion of (a + x)3 is little tougher, but what happens when the expansion is raised to the power of ten or more? So, we have to establish the formula for (a + x)n, where n is any integer. Let us define 'a' as the first term, 'x' as the second term and 'n' as the exponent.

The total terms in the expansion of (a + x)2 and (a + x)3 are 3 and 4 respectively, which means that the number of terms in the expansion is one more than the exponent. So total number of terms in the expansion (a + x)n is (n + 1).

        Now, for n = 2

           (a + x)2 = a2x0 +  2a1x1/1 + (2(2-1)/1*2)a0x2

        = (F.T.)n (S.T.)0 + n/1! (F.T.)(n-1) (S.T.)1 + (n(n-1)/2!) (F.T.)(n-2) (S.T.)2

        = a2 + 2ax + x2

Note: F.T. refers to first term i.e. 'a' and S.T. refers to second term i.e. 'x' 


        (a + x)3 = a3 + 3a2x + ((3(3-1)/(1*2))a0 x2 + ((3(3-1)(3-2))/(1*2*3))x3

                     = a3 + 3a2x + 3ax2 + x3

When n is a positive integer, then

(a + x)n = nC0an + nC1an-1x + nC2an-2 x2 +...+ nCran-r xr +...+ nCnxn,

Where nC0 . nC1 . nC2 ... nCn are called Binomial coefficients.

Proof of Binomial Theorem

Proof of Binomial Theorem is very simple; we can prove it by using the mathematical induction.


Step I:      Let n = 1

                L.H.S. = a + x

                R.H.S. = a + 1C1 x = a + x

        So, theorem is true for n = 1.

Step II:

        Let the theorem be true for n = m, than

        P(m) : (a + x)m = mC0 am + mC1 am-1 x1 + mC2 am-2 +...+ mCm xm       (i)

Step III:

        We have to prove for n = m + 1 i.e. we have to prove that

P(m + 1): (a + x)m+1 = m+1C0 am+1 + m+1C1 am x1

+...+ m+1Cm a xm + m+1Cm+1 xm+1             ....(ii)

        Multiplying by (a + x) on both sides in equation (i), we get,

        (a + x)m+1 = (mC0 am+1 + mC1 am +...+ mCm-1 a2 xm-1 + mCm a xm)

                        = (mC0 am x + mC1 mC0) am x +...+ mCm-1 a xm + mCm xm+1)

Or, (a+x)m+1 = mC0 am+1 + (mC1 + mC0) am x +...+ (mCm-1 + mCm-2) a2 xm-1

+ (mCm + mCm-1) a xm + mCm xm+1

                        = m+1C0 am+1 + m+1C1 am x + m+1C2 am-1 x2 +...+

m+1Cm-1 a2 xm-1 + m+1Cm a xm + m+1Cm+1 xm+1.      

Hence, Proved.


        Expand  (x + 1/x).


         (x + 1/x)7 = 7C0x7 + 7C1x6  (1/x) + 7C2x5 (1/x2) + 7C3x4 + 1/x3 + 7C4x4

(1/x3)  + 7C5x2 + (1/x5) + 7C6x (1/x6) + 7C7 (1/x7)

        = x7 + 7x5 + 21x3 + (35 x) + (35/x )+( 21/x3) + (7/x5)+ (1/x7) .

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