1@akari
\\\texttt{for\: upperbound}\\ 2(\sqrt{k} -\sqrt{k-1}\right))=\frac{2}{\sqrt{k}+\sqrt{k-1}\right}>\frac{1}{\sqrt{k}}\\ \texttt{therefore}\\ \sum_{k=1}^{80}\frac{1}{\sqrt{k}}<1+2\sum_{k=2}^{80}(\sqrt{k}-\sqrt{k-1}\right))=2\sqrt{80}-1<17
(1) If S=\sum_{k=1}^{80}{\frac{1}{\sqrt{k}}} Then \left[S \right]=
where [ .] Greatest Integer function.
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4 Answers
1http://targetiit.com/iit-jee-forum/posts/a-few-ques-from-algebra-12922.html
1i think wat xyz has done is wrong
he has nested
16<S<2√80 =17.somthing
then how can he conclude
[S]=16 , it can be 17 also na ?
so we need to have this inequality
16<=S<17
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