What is expand (e^y+1)^5 ?

The solution to expand (e^y+1)^5 is e^{5y}+5e^{4y}+10e^{3y}+10e^{2y}+5e^y+1

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expand (e^y+1)^5

expand

(e^{y} + 1)^{5}

e^{5 y} + 5 e^{4 y} + 10 e^{3 y} + 10 e^{2 y} + 5 e^{y} + 1

Solution steps

(e^{y} + 1)^{5}

Apply binomial theorem:

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

a = e^{y}, b = 1

= i = 0 \sum 5 (i 5) (e^{y})^{(5 - i)} \cdot 1^{i}

Expand summation

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

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} (e^{y})^{5} \cdot 1^{0} : e^{5 y}

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} (e^{y})^{4} \cdot 1^{1} : 5 e^{4 y}

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} (e^{y})^{3} \cdot 1^{2} : 10 e^{3 y}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} (e^{y})^{2} \cdot 1^{3} : 10 e^{2 y}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} (e^{y})^{1} \cdot 1^{4} : 5 e^{y}

\frac{5 !}{5 ! ( 5 - 5 ) !} (e^{y})^{0} \cdot 1^{5} = 1

= e^{5 y} + 5 e^{4 y} + 10 e^{3 y} + 10 e^{2 y} + 5 e^{y} + 1

36 (4 - π)

\frac{81 ( n ^{2} + n ) ^{2}}{4 n ^{4}}

(\frac{2 i - 2}{n})^{2}

(x - a)^{2} + y^{2}

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

What is expand (e^y+1)^5 ?
The solution to expand (e^y+1)^5 is e^{5y}+5e^{4y}+10e^{3y}+10e^{2y}+5e^y+1