What is expand (b-1/2)^4 ?

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

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expand (b-1/2)^4

expand

(b - \frac{1}{2})^{4}

b^{4} - 2 b^{3} + \frac{3 b ^{2} - b}{2} + \frac{1}{16}

Solution steps

(b - \frac{1}{2})^{4}

Apply binomial theorem:

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

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

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

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

Join

b^{4} - 2 b^{3} + \frac{3 b ^{2}}{2} - \frac{b}{2} + \frac{1}{16} : b^{4} - 2 b^{3} + \frac{3 b ^{2} - b}{2} + \frac{1}{16}

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

(2 + 3) (1 - 2)

3 (x + 1) - 5 ln ∣ x + 1 ∣ + C

\frac{( 2 n + 1 ) ( n + 1 ) n}{6}

2 (2 + 2)

\frac{1}{( s + 1 ) ( s ^{2} + 4 s + 25 )}

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