1. Examples
1. Expand `(x+2)^4`
Solution:
`(x+2)^4`
Using Binomial Theorem,
`(a+b)^n=((n),(0))a^nb^0+((n),(1))a^(n-1)b^1+((n),(2))a^(n-2)b^2+...+((n),(n))a^0b^n`
where `((n),(0))=1,((n),(1))=n,((n),(n))=1,((n),(r))=(n!)/(r!(n-r)!)`
`"Here "a=x,b=2,n=4`
`=((4),(0))(x)^4(2)^0+((4),(1))(x)^3(2)^1+((4),(2))(x)^2(2)^2+((4),(3))(x)^1(2)^3+((4),(4))(x)^0(2)^4`
`=1(x)^4(2)^0+4(x)^3(2)^1+(4!)/(2!(4-2)!)(x)^2(2)^2+(4!)/(3!(4-3)!)(x)^1(2)^3+1(x)^0(2)^4`
`=1(x^4)(1)+4(x^3)(2)+(4*3)/(2*1)(x^2)(4)+(4)/(1)(x)(8)+1(1)(16)`
`=1(x^4)(1)+4(x^3)(2)+6(x^2)(4)+4(x)(8)+1(1)(16)`
`=x^4+8x^3+24x^2+32x+16`
2. Expand `(x-3y)^5`
Solution:
`(x-3y)^5`
Using Binomial Theorem,
`(a+b)^n=((n),(0))a^nb^0+((n),(1))a^(n-1)b^1+((n),(2))a^(n-2)b^2+...+((n),(n))a^0b^n`
where `((n),(0))=1,((n),(1))=n,((n),(n))=1,((n),(r))=(n!)/(r!(n-r)!)`
`"Here "a=x,b=-3y,n=5`
`=((5),(0))(x)^5(-3y)^0+((5),(1))(x)^4(-3y)^1+((5),(2))(x)^3(-3y)^2+((5),(3))(x)^2(-3y)^3+((5),(4))(x)^1(-3y)^4+((5),(5))(x)^0(-3y)^5`
`=1(x)^5(-3y)^0+5(x)^4(-3y)^1+(5!)/(2!(5-2)!)(x)^3(-3y)^2+(5!)/(3!(5-3)!)(x)^2(-3y)^3+(5!)/(4!(5-4)!)(x)^1(-3y)^4+1(x)^0(-3y)^5`
`=1(x^5)(1)+5(x^4)(-3y)+(5*4)/(2*1)(x^3)(9y^2)+(5*4)/(2*1)(x^2)(-27y^3)+(5)/(1)(x)(81y^4)+1(1)(-243y^5)`
`=1(x^5)(1)+5(x^4)(-3y)+10(x^3)(9y^2)+10(x^2)(-27y^3)+5(x)(81y^4)+1(1)(-243y^5)`
`=x^5-15x^4y+90x^3y^2-270x^2y^3+405xy^4-243y^5`
3. Expand `(2x-3)^3`
Solution:
`(2x-3)^3`
`"Using the identity,"`
`(A-B)^3=A^3-B^3-3AB(A-B)`
`"Here "A=2x,B=3`
`=(2x)^3-(3)^3-3(2x)(3)((2x)-(3))`
`=8x^3-27-(18x)(2x-3)`
`=8x^3-27-36x^2+54x`
Second method :
Using Binomial Theorem,
`(a+b)^n=((n),(0))a^nb^0+((n),(1))a^(n-1)b^1+((n),(2))a^(n-2)b^2+...+((n),(n))a^0b^n`
where `((n),(0))=1,((n),(1))=n,((n),(n))=1,((n),(r))=(n!)/(r!(n-r)!)`
`"Here "a=2x,b=-3,n=3`
`=((3),(0))(2x)^3(-3)^0+((3),(1))(2x)^2(-3)^1+((3),(2))(2x)^1(-3)^2+((3),(3))(2x)^0(-3)^3`
`=1(2x)^3(-3)^0+3(2x)^2(-3)^1+(3!)/(2!(3-2)!)(2x)^1(-3)^2+1(2x)^0(-3)^3`
`=1(8x^3)(1)+3(4x^2)(-3)+(3)/(1)(2x)(9)+1(1)(-27)`
`=1(8x^3)(1)+3(4x^2)(-3)+3(2x)(9)+1(1)(-27)`
`=8x^3-36x^2+54x-27`
4. Expand `(3x+y)^4`
Solution:
`(3x+y)^4`
Using Binomial Theorem,
`(a+b)^n=((n),(0))a^nb^0+((n),(1))a^(n-1)b^1+((n),(2))a^(n-2)b^2+...+((n),(n))a^0b^n`
where `((n),(0))=1,((n),(1))=n,((n),(n))=1,((n),(r))=(n!)/(r!(n-r)!)`
`"Here "a=3x,b=y,n=4`
`=((4),(0))(3x)^4(y)^0+((4),(1))(3x)^3(y)^1+((4),(2))(3x)^2(y)^2+((4),(3))(3x)^1(y)^3+((4),(4))(3x)^0(y)^4`
`=1(3x)^4(y)^0+4(3x)^3(y)^1+(4!)/(2!(4-2)!)(3x)^2(y)^2+(4!)/(3!(4-3)!)(3x)^1(y)^3+1(3x)^0(y)^4`
`=1(81x^4)(1)+4(27x^3)(y)+(4*3)/(2*1)(9x^2)(y^2)+(4)/(1)(3x)(y^3)+1(1)(y^4)`
`=1(81x^4)(1)+4(27x^3)(y)+6(9x^2)(y^2)+4(3x)(y^3)+1(1)(y^4)`
`=81x^4+108x^3y+54x^2y^2+12xy^3+y^4`
This material is intended as a summary. Use your textbook for detail explanation.
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