![\displaystyle \sum_{k=0}^{n} a_k=\displaystyle \sum_{k=0}^{n} 7\cdot (\dfrac{4}{9})^k=7\cdot \dfrac{1-(\dfrac{4}{9})^{n+1}}{1-\dfrac{4}{9}}=\dfrac{7*9*[1-(\dfrac{4}{9})^{n+1}]}{9-4}=\dfrac{63}{5}\cdot [1-(\dfrac{4}{9})^{n+1}}]](/tpl/images/0715/4391/2e481.png)
Find the sum of the following infinite geometric series
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