A) \[\frac{5}{4}+\frac{15}{16}\left( 1-\frac{1}{{{5}^{n-1}}} \right)-\frac{(3n-2)}{4\cdot {{5}^{n-1}}}\]
B) \[\frac{5}{4}+\frac{1}{16}\left( 1-\frac{1}{{{5}^{n-1}}} \right)-\frac{3n}{4\cdot {{5}^{n-1}}}\]
C) \[\left( 1-\frac{1}{{{5}^{n-1}}} \right)-\frac{(3n+2)}{4\cdot {{5}^{n-1}}}\]
D) None of the above
Correct Answer: A
Solution :
Clearly, the given series is an arithmetic geometric series whose corresponding AP and GP are respectively,\[1,\,\,4,\,\,7,\,\,10,...\]and\[1,\,\,\frac{1}{5},\,\,\frac{1}{{{5}^{{}}}},\,\,\frac{1}{{{5}^{3}}},...\] The\[{{n}^{th}}\]term of\[AP=[1+(n-1)\times 3]=3n-2\] The\[{{n}^{th}}\]term of\[GP=\left[ 1\times {{\left( \frac{1}{5} \right)}^{n-1}} \right]={{\left( \frac{1}{5} \right)}^{n-1}}\] So, the\[{{n}^{th}}\]term of the given series is \[(3n-2)\times \frac{1}{{{5}^{n-1}}}=\frac{3n-2}{{{5}^{n-1}}}\] Let, \[{{S}_{n}}=1+\frac{4}{5}+\frac{7}{{{5}^{2}}}+\frac{10}{{{5}^{3}}}+...\] \[+\frac{3n-5}{{{5}^{n-2}}}+\frac{3n-2}{{{5}^{n-1}}}\] ? (i) \[\frac{1}{5}{{S}_{n}}=\frac{1}{5}+\frac{4}{{{5}^{2}}}+\frac{7}{{{5}^{3}}}+...+\frac{(3n-5)}{{{5}^{n-1}}}+\frac{3n-2}{{{5}^{n}}}\]? (ii) Subtracting (ii) from (i), we get \[{{S}_{n}}-\frac{1}{5}{{S}_{n}}=1+\left\{ \frac{3}{5}+\frac{3}{{{5}^{2}}}+\frac{3}{{{5}^{3}}}+...+\frac{3}{{{5}^{n-1}}} \right\}\] \[-\frac{(3n-2)}{{{5}^{n}}}\] \[\Rightarrow \]\[\frac{4}{5}{{S}_{n}}=1+\frac{3}{5}\frac{\left\{ 1-{{\left( \frac{1}{5} \right)}^{n-1}} \right\}}{\left( 1-\frac{1}{5} \right)}-\frac{(3n-2)}{{{5}^{n}}}\] \[\Rightarrow \]\[\frac{4}{5}{{S}_{n}}=1+\frac{3}{5}\frac{\left\{ 1-\frac{1}{{{5}^{n-1}}} \right\}}{\left( \frac{4}{5} \right)}-\frac{(3n-2)}{{{5}^{n}}}\] \[\Rightarrow \]\[\frac{4}{5}{{S}_{n}}=1+\frac{3}{4}\left( 1-\frac{1}{{{5}^{n-1}}} \right)-\frac{(3n-2)}{{{5}^{n}}}\] \[\Rightarrow \]\[={{S}_{n}}=\frac{5}{4}+\frac{15}{16}\left( 1-\frac{1}{{{5}^{n-1}}} \right)-\frac{(3n-2)}{4\cdot {{5}^{n-1}}}\]
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