Let $a_1=b_1=1$ and ${a_n} = {a_{n - 1}} + (n - 1),{b_n} = {b_{n - 1}} + {a_{n - 1}},\forall n \ge 2$. If $S = \sum\limits_{n = 1}^{10} {{{{b_n}} \over {{2^n}}}}$ and $T = \sum\limits_{n = 1}^8 {{n \over {{2^{n - 1}}}}}$, then ${2^7}(2S - T)$ is equal to ____________.
Answer (integer)
461
Solution
<p>$\because$ ${a_n} = {a_{n - 1}} + (n - 1)$ and ${a_1} = {b_1} = 1$</p>
<p>${b_n} = {b_{n - 1}} + {a_{n - 1}}$</p>
<p>$\therefore$ ${b_{n + 1}} = 2{b_n} - {b_{n - 1}} + n - 1$</p>
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<p>$\therefore$ $$2S - T = \left( {\sum\limits_{n = 1}^8 {{{{b_n} - n} \over {{2^{n - 1}}}}} } \right) + {{{b_9}} \over {{2^8}}} + {{{b_{10}}} \over {{2^9}}}$$</p>
<p>$= {{461} \over {128}}$</p>
<p>$\therefore$ ${2^7}(2S - T) = 461$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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