(k)/(2k+1)+(1)/((2k+1)(2k+3)) - addition of fractions

(k)/(2k+1)+(1)/((2k+1)(2k+3)) - step by step solution for the given fractions. Addition of fractions, full explanation.

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    Solution for the given fractions

    • k/(2*k+1) + 1/((2*k+1)*(2*k+3)) = ?
    • The common denominator of the two fractions is: (2*k+1)^2*(2*k+3)
    • k/(2*k+1) = (k*(2*k+1)*(2*k+3))/((2*k+1)*(2*k+1)*(2*k+3)) = (k*(2*k+1)*(2*k+3))/((2*k+1)^2*(2*k+3))
    • 1/((2*k+1)*(2*k+3)) = (1*(2*k+1))/((2*k+1)*(2*k+3)*(2*k+1)) = (2*k+1)/((2*k+1)^2*(2*k+3))
    • Fractions adjusted to a common denominator
    • k/(2*k+1) + 1/((2*k+1)*(2*k+3)) = (k*(2*k+1)*(2*k+3))/((2*k+1)^2*(2*k+3)) + (2*k+1)/((2*k+1)^2*(2*k+3))
    • (k*(2*k+1)*(2*k+3))/((2*k+1)^2*(2*k+3)) + (2*k+1)/((2*k+1)^2*(2*k+3)) = (k*(2*k+1)*(2*k+3)+2*k+1)/((2*k+1)^2*(2*k+3))
    • (k*(2*k+1)*(2*k+3)+2*k+1)/((2*k+1)^2*(2*k+3)) = (k*(2*k+1)*(2*k+3)+2*k+1)/((2*k+1)^2*(2*k+3))

    Solution for the given fractions

    $ \frac{k}{(2*k+1)} +\frac{ 1}{((2*k+1)*(2*k+3))} =? $

    The common denominator of the two fractions is: (2*k+1)^2*(2*k+3)

    $ \frac{k}{(2*k+1)} = \frac{(k*(2*k+1)*(2*k+3))}{((2*k+1)*(2*k+1)*(2*k+3))} = \frac{(k*(2*k+1)*(2*k+3))}{((2*k+1)^2*(2*k+3))} $

    $ \frac{1}{((2*k+1)*(2*k+3))} = \frac{(1*(2*k+1))}{((2*k+1)*(2*k+3)*(2*k+1))} = \frac{(2*k+1)}{((2*k+1)^2*(2*k+3))} $

    Fractions adjusted to a common denominator

    $ \frac{k}{(2*k+1)} +\frac{ 1}{((2*k+1)*(2*k+3))} = \frac{(k*(2*k+1)*(2*k+3))}{((2*k+1)^2*(2*k+3))} + \frac{(2*k+1)}{((2*k+1)^2*(2*k+3))} $

    $ \frac{(k*(2*k+1)*(2*k+3))}{((2*k+1)^2*(2*k+3))} + \frac{(2*k+1)}{((2*k+1)^2*(2*k+3))} = \frac{(k*(2*k+1)*(2*k+3)+2*k+1)}{((2*k+1)^2*(2*k+3))} $

    $ \frac{(k*(2*k+1)*(2*k+3)+2*k+1)}{((2*k+1)^2*(2*k+3))} = \frac{(k*(2*k+1)*(2*k+3)+2*k+1)}{((2*k+1)^2*(2*k+3))} $

    $ $

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