How do you find the binomial expansion of #(2x-1)^5#?

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Answer 1 · 2015-08-06T19:33:43

Multiply the #6#th row of Pascal's triangle by a sequence of descending powers of #2# to find the coefficients:

#(2x-1)^5 = 32x^5-80x^4+80x^3-40x^2+10x-1#

Write down the #6#th row of Pascal's triangle as a sequence:

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

Write down descending powers of #2# from #2^5# to #2^0# as a sequence:

#32#, #16#, #8#, #4#, #2#, #1#

Multiply the two sequences together to get:

#32#, #80#, #80#, #40#, #10#, #1#

With suitable alternation of signs, these are the coefficients of the terms in descending powers of #x#:

#(2x-1)^5 = 32x^5-80x^4+80x^3-40x^2+10x-1#