How do you expand the binomial #(x+4)^5# using the binomial theorem?

2 Answers
Pankaj Solanki
Oct 1, 2017

#(x+4)^5=x^5+20x^4+160x^3+640x^2+1280x+1024#

Explanation:

Binomial Expansion
#(x+y)^n= ^nC_0x^ny^0+^nC_1x^(n-1)y^1+^nC_2x^(n-2)y^2+...+^nC_nx^0y^n#
or
#(x+y)^n= nx^ny^0+(n(n-1))/(2!)x^(n-1)y^1+(n(n-1)(n-2))/(3!)x^(n-2)y^2+...+x^0y^n#

Hence
#(x+4)^5=x^5+5^1x^4(4)+(5(4))/(2!)x^3(4^2)+(5(4)(3))/(3!)x^2(4)^3+(5(4)(3)(2))/(4!)x^1(4)^4+(5(4)(3)(2)(1))/(5!)x^0(4)^5#

or

#(x+4)^5=x^5+20x^4+(20)/(2)x^3(4^2)+(60)/(6)x^2(4)^3+(120)/(24)x^1(4)^4+(120)/(120)x^0(4)^5#

or

#(x+4)^5=x^5+20x^4+(10)x^3(16)+(10)/x^2(64)+5x^1(256)+1024#

or

#(x+4)^5=x^5+20x^4+160x^3+640x^2+1280x+1024#

Binayaka C.
Oct 1, 2017

#(x+4)^5 = x^5+20x^4+160x^3+640x^2+1280x+1024#

Explanation:

We know #(a+b)^n= nC_0 a^n*b^0 +nC_1 a^(n-1)*b^1 + nC_2 a^(n-2)*b^2+..........+nC_n a^(n-n)*b^n#

Here #a=x,b=4,n=5# We know, #nC_r = (n!)/(r!*(n-r)!#

#:.5C_0 =1 , 5C_1 =5, 5C_2 =10,5C_3 =10, 5C_4 =5,5C_5 =1#

#(x+4)^5 = x^5+5*x^4*4+10*x^3*4^2+10*x^2*4^3+5*x*4^4+4^5# or

#(x+4)^5 = x^5+20x^4+160x^3+640x^2+1280x+1024# [Ans]

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