What is expand (1+e^{2x})^4 ?

The solution to expand (1+e^{2x})^4 is 1+4e^{2x}+6e^{4x}+4e^{6x}+e^{8x}

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expand (1+e^{2x})^4

expand

(1 + e^{2 x})^{4}

1 + 4 e^{2 x} + 6 e^{4 x} + 4 e^{6 x} + e^{8 x}

Solution steps

(1 + e^{2 x})^{4}

Apply binomial theorem:

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

a = 1, b = e^{2 x}

= i = 0 \sum 4 (i 4) \cdot 1^{(4 - i)} (e^{2 x})^{i}

Expand summation

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

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

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 1^{3} (e^{2 x})^{1} : 4 e^{2 x}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 1^{2} (e^{2 x})^{2} : 6 e^{4 x}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 1^{1} (e^{2 x})^{3} : 4 e^{6 x}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 1^{0} (e^{2 x})^{4} : e^{8 x}

= 1 + 4 e^{2 x} + 6 e^{4 x} + 4 e^{6 x} + e^{8 x}

\frac{( 2 \cdot π \cdot s )}{( s ^{2} + π ^{2} ) ^{2}}

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

(- 4, 4) y (2, - 8)

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

\frac{z ^{2} - z + 0.64}{( z - 1 ) ( z - 0.36 )} - 1

What is expand (1+e^{2x})^4 ?
The solution to expand (1+e^{2x})^4 is 1+4e^{2x}+6e^{4x}+4e^{6x}+e^{8x}