What is expand (x/9+1)^4 ?

The solution to expand (x/9+1)^4 is (x^4}{6561}+(4x^3)/(729)+(2x^2)/(27)+\frac{4x)/9+1

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expand (x/9+1)^4

expand

(\frac{x}{9} + 1)^{4}

\frac{x ^{4}}{6561} + \frac{4 x ^{3}}{729} + \frac{2 x ^{2}}{27} + \frac{4 x}{9} + 1

Solution steps

(\frac{x}{9} + 1)^{4}

Apply binomial theorem:

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

a = \frac{x}{9}, b = 1

= i = 0 \sum 4 (i 4) (\frac{x}{9})^{(4 - i)} \cdot 1^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} (\frac{x}{9})^{4} \cdot 1^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (\frac{x}{9})^{3} \cdot 1^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (\frac{x}{9})^{2} \cdot 1^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (\frac{x}{9})^{1} \cdot 1^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (\frac{x}{9})^{0} \cdot 1^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (\frac{x}{9})^{4} \cdot 1^{0} : \frac{x ^{4}}{6561}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (\frac{x}{9})^{3} \cdot 1^{1} : \frac{4 x ^{3}}{729}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (\frac{x}{9})^{2} \cdot 1^{2} : \frac{2 x ^{2}}{27}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (\frac{x}{9})^{1} \cdot 1^{3} : \frac{4 x}{9}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (\frac{x}{9})^{0} \cdot 1^{4} : 1

= \frac{x ^{4}}{6561} + \frac{4 x ^{3}}{729} + \frac{2 x ^{2}}{27} + \frac{4 x}{9} + 1

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What is expand (x/9+1)^4 ?
The solution to expand (x/9+1)^4 is (x^4}{6561}+(4x^3)/(729)+(2x^2)/(27)+\frac{4x)/9+1