If the value of $\left(1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\ldots \text{ to } \infty\right)^{\log_{(0.25)}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots \text{ to } \infty\right)}$ is $l$,then $l^{2}$ is equal to $......$

  • A
    $2$
  • B
    $3$
  • C
    $4$
  • D
    $5$

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