Canada
If there are ^@ n ^@ numbers of which one is ^@ \left( 1 - \dfrac{ 1 } { n^5 } \right) ^@ and all the others are ^@ 1's, ^@ then by how much is the arithmetic mean of these numbers less than ^@1.^@
Answer:
^@ \dfrac{ 1 } { n^6 } ^@
- It is given that there are ^@ n ^@ numbers of which one is ^@ \left( 1 - \dfrac{ 1 } { n^5 } \right) ^@ and all the others are ^@ 1's. ^@
Therefore, the numbers are ^@ \left( 1 - \dfrac{ 1 } { n^5 } \right), 1, 1, 1 \ldots ^@ (where ^@ n ^@ is the total number of numbers in the series) - Out of ^@ n ^@ numbers one is ^@ \left( 1 - \dfrac{ 1 } { n^5 } \right) ^@ and remaining ^@n-1^@ numbers are ^@ 1. ^@
Therefore, the sum of ^@ n-1 ^@ numbers is ^@ n-1.^@
Now, the sum of all numbers in the series ^@ = n-1 + \left( 1 - \dfrac{ 1 } { n^5 } \right) = n - \dfrac{ 1 } { n^5 } ^@ - Now, the arithmetic mean of the numbers ^@ = ^@
Sum of the all numbers ^@ n ^@
^@ = \dfrac { n - \dfrac{ 1 } { n^5 } } { n } ^@
^@ = \dfrac{ n } { n } - \dfrac{ 1 } { n^6 } ^@
^@ = 1 - \dfrac{ 1 } { n^6 } ^@ - Thus, we can say that the arithmetic mean of these numbers is ^@ \dfrac{ 1 } { n^6 } ^@ less than ^@1.^@
