What is expand (e^x+2)^5 ?

The solution to expand (e^x+2)^5 is e^{5x}+10e^{4x}+40e^{3x}+80e^{2x}+80e^x+32

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expand (e^x+2)^5

expand

(e^{x} + 2)^{5}

e^{5 x} + 10 e^{4 x} + 40 e^{3 x} + 80 e^{2 x} + 80 e^{x} + 32

Solution steps

(e^{x} + 2)^{5}

Apply binomial theorem:

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

a = e^{x}, b = 2

= i = 0 \sum 5 (i 5) (e^{x})^{(5 - i)} \cdot 2^{i}

Expand summation

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

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} (e^{x})^{5} \cdot 2^{0} : e^{5 x}

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} (e^{x})^{4} \cdot 2^{1} : 10 e^{4 x}

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} (e^{x})^{3} \cdot 2^{2} : 40 e^{3 x}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} (e^{x})^{2} \cdot 2^{3} : 80 e^{2 x}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} (e^{x})^{1} \cdot 2^{4} : 80 e^{x}

\frac{5 !}{5 ! ( 5 - 5 ) !} (e^{x})^{0} \cdot 2^{5} = 32

= e^{5 x} + 10 e^{4 x} + 40 e^{3 x} + 80 e^{2 x} + 80 e^{x} + 32

\frac{2}{( ( x ^{2} + 1 ) \cdot x ^{2} )}

\frac{4 x - 2}{x ( x + 1 ) ( x - 2 )}

\frac{2 s ^{2} + 4 s + 5}{s ( s + 1 )}

\frac{1}{( x - 1 ) ( x + 3 )}

(c + Δ x)^{2}

What is expand (e^x+2)^5 ?
The solution to expand (e^x+2)^5 is e^{5x}+10e^{4x}+40e^{3x}+80e^{2x}+80e^x+32