What is expand (e^x+1)^7 ?

The solution to expand (e^x+1)^7 is e^{7x}+7e^{6x}+21e^{5x}+35e^{4x}+35e^{3x}+21e^{2x}+7e^x+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand (e^x+1)^7

expand

(e^{x} + 1)^{7}

e^{7 x} + 7 e^{6 x} + 21 e^{5 x} + 35 e^{4 x} + 35 e^{3 x} + 21 e^{2 x} + 7 e^{x} + 1

Solution steps

(e^{x} + 1)^{7}

Apply binomial theorem:

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

a = e^{x}, b = 1

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

Expand summation

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

Simplify

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

Simplify

\frac{7 !}{1 ! ( 7 - 1 ) !} (e^{x})^{6} \cdot 1^{1} : 7 e^{6 x}

Simplify

\frac{7 !}{2 ! ( 7 - 2 ) !} (e^{x})^{5} \cdot 1^{2} : 21 e^{5 x}

Simplify

\frac{7 !}{3 ! ( 7 - 3 ) !} (e^{x})^{4} \cdot 1^{3} : 35 e^{4 x}

Simplify

\frac{7 !}{4 ! ( 7 - 4 ) !} (e^{x})^{3} \cdot 1^{4} : 35 e^{3 x}

Simplify

\frac{7 !}{5 ! ( 7 - 5 ) !} (e^{x})^{2} \cdot 1^{5} : 21 e^{2 x}

Simplify

\frac{7 !}{6 ! ( 7 - 6 ) !} (e^{x})^{1} \cdot 1^{6} : 7 e^{x}

\frac{7 !}{7 ! ( 7 - 7 ) !} (e^{x})^{0} \cdot 1^{7} = 1

= e^{7 x} + 7 e^{6 x} + 21 e^{5 x} + 35 e^{4 x} + 35 e^{3 x} + 21 e^{2 x} + 7 e^{x} + 1

8 y^{^{'}} + t y = 7

\frac{1}{x - 1}

\int \frac{8 z ^{2} + 2}{2 z ^{3} + z}

\frac{d}{d x} (\frac{e ^{x} cos ( x )}{e ^{x} - 1})

What is expand (e^x+1)^7 ?
The solution to expand (e^x+1)^7 is e^{7x}+7e^{6x}+21e^{5x}+35e^{4x}+35e^{3x}+21e^{2x}+7e^x+1