What is (k!(-2)^2k)/((3^2k+1)) ?

The solution to (k!(-2)^2k)/((3^2k+1)) is (4k!k)/(9k+1)

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simplify (k!(-2)^2k)/((3^2k+1))

simplify

\frac{k ! ( - 2 ) ^{2} k}{( 3 ^{2} k + 1 )}

\frac{4 k ! k}{9 k + 1}

Solution steps

\frac{k ! ( - 2 ) ^{2} k}{3 ^{2} k + 1}

(- 2)^{2} = 4

= \frac{k ! \cdot 4 k}{3 ^{2} k + 1}

3^{2} = 9

= \frac{k ! \cdot 4 k}{9 k + 1}

= \frac{4 k ! k}{9 k + 1}

p^{4} + 2 p^{2} + 9

2 x^{4} + 54 x

(5 a^{4} b^{3}) (- 2 a^{2} b^{5}) (4 a^{2} b)

(- 4 x^{4} \cdot 4 y^{3} \cdot z)^{2}

(\frac{( 16 w ^{8} )}{( y ^{20} )})^{\frac{- 3}{4}}

What is (k!(-2)^2k)/((3^2k+1)) ?
The solution to (k!(-2)^2k)/((3^2k+1)) is (4k!k)/(9k+1)