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Question: Answered & Verified by Expert
Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots \ldots . \mathrm{a}_{100}$ be non-zero real numbers such that $\mathrm{a}_{1}+\mathrm{a}_{2}+\ldots \ldots .+\mathrm{a}_{100}=0$, Then
MathematicsSequences and SeriesKVPYKVPY 2016 (SA)
Options:
  • A $\sum_{j=1}^{100} a_{i} 2^{a_{i}}>0$ and $\sum_{i=1}^{100} a_{i} 2^{-a_{i}} < 0$
  • B $\sum_{\mathrm{i}=1}^{100} \mathrm{a}_{\mathrm{i}} 2^{\mathrm{a}_{i}} \geq 0$ and $\sum_{\mathrm{i}=1}^{100} \mathrm{a}_{\mathrm{i}} 2^{-\mathrm{a}_{\mathrm{i}}} \geq 0$
  • C $\sum_{i=1}^{100} a_{i} 2^{a_{i}} \leq 0$ and $\sum_{i=1}^{100} a_{i} 2^{-a_{i}} \leq 0$
  • D the sign of $\sum_{i=1}^{100} a_{i} 2^{a_{i}}$ or $\sum_{j=1}^{100} a_{i} 2^{-a_{i}}$ depends on the choice of $a_{i}$ 's
Solution:
2944 Upvotes Verified Answer
The correct answer is: $\sum_{j=1}^{100} a_{i} 2^{a_{i}}>0$ and $\sum_{i=1}^{100} a_{i} 2^{-a_{i}} < 0$
Note that for every real number $\mathrm{a}_{\mathrm{i}}$
$\begin{array}{ll}
& a_{i} \cdot 2^{a_{i}}>a_{i} \quad \text { and } \quad a_{i} \cdot 2^{-a_{i}} < a_{i} \\
\text { therefore, } & \sum_{i=1}^{100} a_{i} \cdot 2^{a_{i}}>\sum_{i=1}^{100} a_{i} \text { and } \sum_{i=1}^{100} a_{i} \cdot 2^{-a_{i}}>\sum_{i=1}^{100} a_{i}
\end{array}$

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