Given that \(\displaystyle \sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ 2 } } } \approx 1.645\),

If \(X=\frac { 1 }{ { 1 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 2 } } +...+\frac { 1 }{ { 100 }^{ 2 } } \), find the first 2 digits after the decimal point of X. (Note: the 'first 2 digits' is without rounding.)

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