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Please evaluate\(1/1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+...+2018)\)

Best Answer 

 #1
avatar+394 
+2

\(\frac{1}{\frac{1(1+1)}{2}}+\frac{1}{\frac{2(2+1)}{2}}+\dots + \frac{1}{\frac{2018(2018+1)}{2}}\)

\(\frac{2}{1(2)}+\frac{2}{2(3)}+ \dots+\frac{2}{2018(2019)}\)

\(2(\frac{1}{1(2)}+ \frac{1}{2(3)} + \dots + \frac{1}{2018(2019)})\)
\(\text{Apply } \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\)

\(2(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3} +\dots -\frac{1}{2019})\)

\(2(1-\frac{1}{2019})\)

\(\frac{4036}{2019}\)

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 Apr 4, 2024
 #1
avatar+394 
+2
Best Answer

\(\frac{1}{\frac{1(1+1)}{2}}+\frac{1}{\frac{2(2+1)}{2}}+\dots + \frac{1}{\frac{2018(2018+1)}{2}}\)

\(\frac{2}{1(2)}+\frac{2}{2(3)}+ \dots+\frac{2}{2018(2019)}\)

\(2(\frac{1}{1(2)}+ \frac{1}{2(3)} + \dots + \frac{1}{2018(2019)})\)
\(\text{Apply } \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\)

\(2(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3} +\dots -\frac{1}{2019})\)

\(2(1-\frac{1}{2019})\)

\(\frac{4036}{2019}\)

hairyberry Apr 4, 2024

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