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 $x$ :

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

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