Monthly Archives: ខែកក្កដា 2009

គណិតវិទ្យាជុំវិញពិភពលោក(៣)

Posted on ខែកក្កដា 28, 2009 | មតិ 29

ចូរបង្ហាញថា\sum\limits_{n = 1}^{9999} {\left( {{1 \over {\left( {\sqrt n + \sqrt {n+ 1} } \right)\left( {\root 4 \of n + \root 4 \of {n + 1} } \right)}}} \right)} = 9។

ដំណោះស្រាយ

បង្ហាញថា\sum\limits_{n = 1}^{9999} {\left( {{1 \over {\left( {\sqrt n + \sqrt {n+ 1} } \right)\left( {\root 4 \of n + \root 4 \of {n + 1} } \right)}}} \right)} = 9

គេមាន\left( {\root 4 \of {n + 1} + \root 4 \of n } \right)\left( {\root 4\of {n + 1} - \root 4 \of n } \right) = \sqrt {n + 1} - \sqrt n

\left( {\root 4 \of {n + 1} + \root 4 \of n } \right)\left( {\root 4\of {n +1} - \root 4 \of n } \right) = {1 \over {\sqrt {n + 1} + \sqrt n }}

គេទាញ{1 \over {\left( {\sqrt n + \sqrt {n + 1} } \right)\left( {\root 4\of n + \root 4 \of {n + 1} } \right)}} = \root 4\of {n + 1} - \root 4 \of n

\sum\limits_{n = 1}^{9999} {\left( {{1 \over {\left( {\sqrt n + \sqrt {n + 1} } \right)\left( {\root 4\of n + \root 4\of {n + 1} } \right)}}} \right) = \sum\limits_{n = 1}^{9999} {\left( {\root 4\of {n + 1} - 1} \right)} }

\sum\limits_{n = 1}^{9999} {\left( {\root 4\of {n + 1} - \root 4\of n } \right) = \root 4\of {10000} - 1 = 9}

ដូចនេះ\sum\limits_{n = 1}^{9999} {\left( {{1 \over {\left( {\sqrt n + \sqrt {n+ 1} } \right)\left( {\root 4 \of n + \root 4 \of {n + 1} } \right)}}} \right)} = 9។

29 មតិ

បានផ្សាយ​ក្នុង គណិតវិទ្យា, ពិជគណិត