### Compound interest and sum of geometric sequence

Question Sample Titled 'Compound interest and sum of geometric sequence'

Nancy invests $${1750}$ at the beginning of 2011 at an interest rate of ${4}\%$ per annum, compounded yearly and she is going to invest the same amount$${1750}$ more at the beginning of every year starting from 2012. At the end of 2032, what is the interest she can earn correct to the nearsest dollar?

A
$${23831}$ B $${19684}$
C
$${58184}$ D $${62331}$

 Let ${I}=$$${1750}$ and ${R}={1}+{4}\%.$ At the end of 2011, she can get  $={I}{R}$ At the end of 2012, she can get  $={\left({I}{R}+{I}\right)}{R}$  $={I}{\left({R}^{{2}}+{R}\right)}$ At the end of 2013, she can get  $={\left({I}{R}^{{2}}+{I}{R}\right)}{R}$  $={I}{\left({R}^{{3}}+{R}^{{2}}+{R}\right)}$ $\ldots$ At the end of 2032, she can get  $={I}{\left({R}^{{22}}+{R}^{{21}}+\ldots+{R}^{{2}}+{R}\right)}$  $={I}{\left(\dfrac{{{R}{\left({R}^{{22}}-{1}\right)}}}{{{R}-{1}}}\right)}$  $\therefore$ The interest that Nancy can earn $={I}{\left(\dfrac{{{R}{\left({R}^{{22}}-{1}\right)}}}{{{R}-{1}}}\right)}-{22}\cdot{1750}$  $=$$${23831.30501347508}$

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