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Binomial Theorem for Positive Integral Indices

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Binomial Theorem for Positive Integral Indices - Lesson Summary

Pascal's Triangle

Consider the identities,

(a+b)0 = 1

(a+b)1 = a+b

(a+b)2 = a2 + 2ab + b2

(a+b)3 = a3 + 3a2b+3ab2 + b3

(a+b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

If the index of the binomial is 0, then the number of terms in the expansion is 1. If the index is 1, then there are two terms in the expansion.

The powers of the first quantity, 'a,' decrease by 1, whereas the powers of the second quantity, 'b,' increase by 1, in each successive term.

In every term, the sum of the powers of a and b is equal to the power of the binomial.

The coefficients of the first and the last terms of an expansion are equal to 1.

The number of terms in the expansion of a binomial is one more than the power of the binomial.

The structure of Pascal triangle looks like a triangle which was developed by a French mathematician, Blaise Pascal.

It is known as Pascal's Triangle.

Binomial Theorem

Concept of combinations is used to write the coefficients when the power of a binomial is very big.

nCr = n!/((n-r)!r!), 0 ≤ r ≤ n

Where n is a non-negative integer.

The binomial theorem for positive integral index states that:

(x+y)n = nC0xn + nC1xn-1y + nC2xn-2y2 + ......... nCn-1xyn-1 + nCnyn

The coefficients of the expansion nC0, nC1, nC2, ... nCn-1, nCn are called binomial coefficients.

nC0 = 1 = nCn

Pascal's triangle using combinations:

Proof of binomial theorem using mathematical induction:

P(n): (x + y)n = nC0 xn + nC1 xn-1 y + nC2 xn-2 y2 + .........nCn-1 x yn-1 + nCn yn

For n = 1

P(1): (x + y)1 = 1C0 x1 + 1C1 y1 = x + y, which is true.

Let us suppose that P(k) is true for some positive integer k.

Thus, we get,

P(k): (x + y)k = kC0 xk + kC1 xk-1 y + kC2 xk-2 y2 + .........kCk-1xyk-1 + kCkyk...............(1)

P(k+1) : (x+y)k+1 = (k+1)C0xk+1 +( k+1)C1xky + (k+1)C2xk-1y2 + ......... + (k+1)Ck+1yk+1

(x+y)k+1

(x+y)k+1 = (x+y) (x+y)k

= (x+y) (kC0xk + kC1xk-1y + kC2xk-2y2 + .........kCk-1xyk-1 + kCkyk)

= kC0xk+1 + kC1xky + kC2xk-1y2 + .........kCk-1x2yk-1 + kCkxyk + kC0xky + kC1xk-1y2 + kC2xk-2y3 + ......... kCk-1xyk + kCkyk+1

Grouping the like terms, we get

= kC0xk+1 +(kC1 + kC0)xky + (kC2 + kC1 )xk-1y2 + .................. (kC1 +kCk-1 )xyk + kCkyk+1

kC0 =1 = k+1C0; kCr + kCr-1 = k+1Cr, and kCk = k+1 Ck+1 = 1

⇒ (x+y)k+1 = k+1C0xk+1 + (k+1)C1xky + (k+1)C2xk-1y2 + .................. (k+1)Ck xyk + (k+1)Ck+1yk+1

P (k+1) is true whenever P(k) is true.

Hence, P(n) is true for all positive integral values of n.

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