The greatest integer less than or equal to the sum of first 100 terms of the sequence ${1 \over 3},{5 \over 9},{{19} \over {27}},{{65} \over {81}},$ ...... is equal to ___________.
Answer (integer)
98
Solution
<p>$S = {1 \over 3} + {5 \over 9} + {{19} \over {27}} + {{65} \over {81}}\, +$ ....</p>
<p>$= \sum\limits_{r = 1}^{100} {\left( {{{{3^r} - {2^r}} \over {{3^r}}}} \right)}$</p>
<p>$$ = 100 - {2 \over 3}{{\left( {1 - {{\left( {{2 \over 3}} \right)}^{100}}} \right)} \over {1/3}}$$</p>
<p>$= 98 + 2{\left( {{2 \over 3}} \right)^{100}}$</p>
<p>$\therefore$ $[S] = 98$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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