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

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Answer 1 · 2015-08-06T19:33:43

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

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

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

$1$ $1$ , $5$ $5$ , $10$ $10$ , $10$ $10$ , $5$ $5$ , $1$ $1$

Write down descending powers of $2$ $2$ from $2^{5}$ $2^5$ to $2^{0}$ $2^0$ as a sequence:

$32$ $32$ , $16$ $16$ , $8$ $8$ , $4$ $4$ , $2$ $2$ , $1$ $1$

Multiply the two sequences together to get:

$32$ $32$ , $80$ $80$ , $80$ $80$ , $40$ $40$ , $10$ $10$ , $1$ $1$

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

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

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