How do you expand #(x + 2)^5# using Pascal’s Triangle?

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Answer 1 · 2015-12-31T07:07:18

#x^5+10x^4+40x^3+80x^2+80x+32#

The #5"th"# row of Pascal's triangle is

#1,5,10,10,5,1#

These values are the coefficients in a binomial expansion to the #5"th"# power.

#(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5#

Notice the pattern of the exponents: the exponent of #a# starts at #5# and goes to #0#, and #b# starts at #0# and increases to #5#.

Apply the rule to #(x+2)#:

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

#=>x^5+10x^4+40x^3+80x^2+80x+32#