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Special Cases of Binomial Theorem

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Special Cases of Binomial Theorem - Lesson Summary

A binomial raised to any positive integral index can be expanded using Pascal's triangle.

Expansion of a binomial using the Binomials theorem:

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

Special case (i)

When x = a and y = - b

(a + (-b))n = nC0 an + nC1 an-1 (-b) + nC2 an-2 (-b)2 + ......... nCn-1 a (-b)n-1 + nCn (-b)n

(a - b)n = nC0 an + nC1 an-1 (-b) + nC2 an-2 (-b)2 + ......... nCn-1 a (-b)n-1 + nCn (-b)n

(a - b)n = nC0 an - nC1 an-1 b + nC2 an-2 b2 + ......... nCn-1 a (-b)n-1 + nCn bn (-1)n

Ex: Expand (2x - y)4

Solution:

(2x - y)4 = 4C0 (2x)4 - 4C1 (2x)4-1 y + 4C2 (2x)4-2 y2 -4C3 (2x)4-3 y3 + 4C4 y4

(2x - y)4 = (2x)4 - 4 (2x)3 y + 6 (2x)2 y2 -4 (2x) y3 + y4

(2x - y)4 = 16 x4 - 32 x3y + 24 x2y2 - 8 xy3 + y4

Special Case (ii)

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

When x = 1

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

(1 + y)n = nC0 + nC1 y + nC2 y2 + ......... nCn-1 yn-1 + nCn yn

When y = 1

(1+ 1)n = nC0 + nC1 (1) + nC2 (1)2 + ......... nCn-1 (1)n-1 + nCn (1)n

2n = nC0 + nC1 + nC2 + .........nCn-1 + nCn

This is a formula to find the sum of the coefficients of a binomial expansion.

Special case (iii)

(1+x)n = nC0 (1)n + nC1 (1)n-1 x + nC2 (1)n-2 x2 + .........nCn-1 (1) (x)n-1 + nCn (x)n

When x = -y

Þ (1 + (-y))n = nC0 (1)n + nC1 (1)n-1(-y) + nC2 (1)n-2(-y)2 + .........nCn-1 (1)(-y)n-1 + nCn (-y)n

Þ (1 - y)n = nC0 - nC1 y + nC2 y2 - nC3 y3 + ..............+(-1)n nCn yn

When y = 1

Þ (1 - 1)n = nC0 - nC1 + nC2 - nC3 + .........+(-1)n nCn

Þ nC0 - nC1 + nC2 - nC3 + .........+(-1)n nCn= 0

Þ nC0 + nC2 + nC4 + ......... = nC1 + nC3 + nC5 ..........

Sum of the coefficients of odd numbered terms = Sum of the coefficients of even numbered terms

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