What is expand (1+1/2 x)^4 ?

The solution to expand (1+1/2 x)^4 is 1+2x+(3x^2)/2+(x^3)/2+(x^4)/(16)

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expand (1+1/2 x)^4

expand

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

1 + 2 x + \frac{3 x ^{2}}{2} + \frac{x ^{3}}{2} + \frac{x ^{4}}{16}

Solution steps

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

Apply binomial theorem:

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

a = 1, b = \frac{1}{2} x

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 1^{0} (\frac{1}{2} x)^{4} : \frac{x ^{4}}{16}

= 1 + 2 x + \frac{3 x ^{2}}{2} + \frac{x ^{3}}{2} + \frac{x ^{4}}{16}

(4 x - x^{2}) cos (y)

, (x - 2) 44 x - x^{2} d x

\frac{3}{( p + 2 ) ( p + 3 ) ( p - 1 + j \cdot 3 ) ( p - 1 - j \cdot 3 )}

\frac{x ( 5 - x )}{( 2 + x )}

\frac{s}{( s ^{2} + 9 ) ( s + 3 ) ^{2}}

What is expand (1+1/2 x)^4 ?
The solution to expand (1+1/2 x)^4 is 1+2x+(3x^2)/2+(x^3)/2+(x^4)/(16)