What is expand (2x-1)^7 ?

The solution to expand (2x-1)^7 is 128x^7-448x^6+672x^5-560x^4+280x^3-84x^2+14x-1

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expand (2x-1)^7

expand

(2 x - 1)^{7}

128 x^{7} - 448 x^{6} + 672 x^{5} - 560 x^{4} + 280 x^{3} - 84 x^{2} + 14 x - 1

Solution steps

(2 x - 1)^{7}

Apply binomial theorem:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = 2 x, b = - 1

= i = 0 \sum 7 (i 7) (2 x)^{(7 - i)} (- 1)^{i}

Expand summation

= \frac{7 !}{0 ! ( 7 - 0 ) !} (2 x)^{7} (- 1)^{0} + \frac{7 !}{1 ! ( 7 - 1 ) !} (2 x)^{6} (- 1)^{1} + \frac{7 !}{2 ! ( 7 - 2 ) !} (2 x)^{5} (- 1)^{2} + \frac{7 !}{3 ! ( 7 - 3 ) !} (2 x)^{4} (- 1)^{3} + \frac{7 !}{4 ! ( 7 - 4 ) !} (2 x)^{3} (- 1)^{4} + \frac{7 !}{5 ! ( 7 - 5 ) !} (2 x)^{2} (- 1)^{5} + \frac{7 !}{6 ! ( 7 - 6 ) !} (2 x)^{1} (- 1)^{6} + \frac{7 !}{7 ! ( 7 - 7 ) !} (2 x)^{0} (- 1)^{7}

Simplify

\frac{7 !}{0 ! ( 7 - 0 ) !} (2 x)^{7} (- 1)^{0} : 128 x^{7}

Simplify

\frac{7 !}{1 ! ( 7 - 1 ) !} (2 x)^{6} (- 1)^{1} : - 448 x^{6}

Simplify

\frac{7 !}{2 ! ( 7 - 2 ) !} (2 x)^{5} (- 1)^{2} : 672 x^{5}

Simplify

\frac{7 !}{3 ! ( 7 - 3 ) !} (2 x)^{4} (- 1)^{3} : - 560 x^{4}

Simplify

\frac{7 !}{4 ! ( 7 - 4 ) !} (2 x)^{3} (- 1)^{4} : 280 x^{3}

Simplify

\frac{7 !}{5 ! ( 7 - 5 ) !} (2 x)^{2} (- 1)^{5} : - 84 x^{2}

Simplify

\frac{7 !}{6 ! ( 7 - 6 ) !} (2 x)^{1} (- 1)^{6} : 14 x

Simplify

\frac{7 !}{7 ! ( 7 - 7 ) !} (2 x)^{0} (- 1)^{7} : - 1

= 128 x^{7} - 448 x^{6} + 672 x^{5} - 560 x^{4} + 280 x^{3} - 84 x^{2} + 14 x - 1

(12.6 + \frac{1}{3}) \cdot (- 0.4)

5 m^{0}

(2 - 3 i) + (6 + 8 i)

- 2 \div \frac{1}{3}

What is expand (2x-1)^7 ?
The solution to expand (2x-1)^7 is 128x^7-448x^6+672x^5-560x^4+280x^3-84x^2+14x-1