What is expand (sqrt(2z)+1)^{12} ?

The solution to expand (sqrt(2z)+1)^{12} is

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expand (sqrt(2z)+1)^{12}

expand

(2 z + 1)^{12}

64 z^{6} + 3842 z^{5} z + 2112 z^{5} + 35202 z^{4} z + 7920 z^{4} + 63362 z^{3} z + 7392 z^{3} + 31682 z^{2} z + 1980 z^{2} + 4402 z z + 132 z + 122 z + 1

Solution steps

(2 z + 1)^{12}

Apply binomial theorem:

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

a = 2 z, b = 1

= i = 0 \sum 12 (i 12) (2 z)^{(12 - i)} \cdot 1^{i}

Expand summation

= \frac{12 !}{0 ! ( 12 - 0 ) !} (2 z)^{12} \cdot 1^{0} + \frac{12 !}{1 ! ( 12 - 1 ) !} (2 z)^{11} \cdot 1^{1} + \frac{12 !}{2 ! ( 12 - 2 ) !} (2 z)^{10} \cdot 1^{2} + \frac{12 !}{3 ! ( 12 - 3 ) !} (2 z)^{9} \cdot 1^{3} + \frac{12 !}{4 ! ( 12 - 4 ) !} (2 z)^{8} \cdot 1^{4} + \frac{12 !}{5 ! ( 12 - 5 ) !} (2 z)^{7} \cdot 1^{5} + \frac{12 !}{6 ! ( 12 - 6 ) !} (2 z)^{6} \cdot 1^{6} + \frac{12 !}{7 ! ( 12 - 7 ) !} (2 z)^{5} \cdot 1^{7} + \frac{12 !}{8 ! ( 12 - 8 ) !} (2 z)^{4} \cdot 1^{8} + \frac{12 !}{9 ! ( 12 - 9 ) !} (2 z)^{3} \cdot 1^{9} + \frac{12 !}{10 ! ( 12 - 10 ) !} (2 z)^{2} \cdot 1^{10} + \frac{12 !}{11 ! ( 12 - 11 ) !} (2 z)^{1} \cdot 1^{11} + \frac{12 !}{12 ! ( 12 - 12 ) !} (2 z)^{0} \cdot 1^{12}

Simplify

\frac{12 !}{0 ! ( 12 - 0 ) !} (2 z)^{12} \cdot 1^{0} : 64 z^{6}

Simplify

\frac{12 !}{1 ! ( 12 - 1 ) !} (2 z)^{11} \cdot 1^{1} : 3842 (z)^{11}

Simplify

\frac{12 !}{2 ! ( 12 - 2 ) !} (2 z)^{10} \cdot 1^{2} : 2112 z^{5}

Simplify

\frac{12 !}{3 ! ( 12 - 3 ) !} (2 z)^{9} \cdot 1^{3} : 35202 (z)^{9}

Simplify

\frac{12 !}{4 ! ( 12 - 4 ) !} (2 z)^{8} \cdot 1^{4} : 7920 z^{4}

Simplify

\frac{12 !}{5 ! ( 12 - 5 ) !} (2 z)^{7} \cdot 1^{5} : 63362 (z)^{7}

Simplify

\frac{12 !}{6 ! ( 12 - 6 ) !} (2 z)^{6} \cdot 1^{6} : 7392 z^{3}

Simplify

\frac{12 !}{7 ! ( 12 - 7 ) !} (2 z)^{5} \cdot 1^{7} : 31682 (z)^{5}

Simplify

\frac{12 !}{8 ! ( 12 - 8 ) !} (2 z)^{4} \cdot 1^{8} : 1980 z^{2}

Simplify

\frac{12 !}{9 ! ( 12 - 9 ) !} (2 z)^{3} \cdot 1^{9} : 4402 (z)^{3}

Simplify

\frac{12 !}{10 ! ( 12 - 10 ) !} (2 z)^{2} \cdot 1^{10} : 132 z

Simplify

\frac{12 !}{11 ! ( 12 - 11 ) !} (2 z)^{1} \cdot 1^{11} : 122 z

Simplify

\frac{12 !}{12 ! ( 12 - 12 ) !} (2 z)^{0} \cdot 1^{12} : 1

= 64 z^{6} + 3842 (z)^{11} + 2112 z^{5} + 35202 (z)^{9} + 7920 z^{4} + 63362 (z)^{7} + 7392 z^{3} + 31682 (z)^{5} + 1980 z^{2} + 4402 (z)^{3} + 132 z + 122 z + 1

(z)^{11} = z^{5} z

(z)^{9} = z^{4} z

(z)^{7} = z^{3} z

(z)^{5} = z^{2} z

(z)^{3} = z z

= 64 z^{6} + 3842 z^{5} z + 2112 z^{5} + 35202 z^{4} z + 7920 z^{4} + 63362 z^{3} z + 7392 z^{3} + 31682 z^{2} z + 1980 z^{2} + 4402 z z + 132 z + 122 z + 1

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(2 x + y = 9,) (2 x - y = 11)

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