Question 990264
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Show that n <= 1 + {{{sqrt(2)}}} + {{{sqrt(3)}}} + . . . + {{{sqrt(n)}}} <= {{{(n(n+1))/2}}}
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It is very easy.


First, &nbsp;n <= 1 + {{{sqrt(2)}}} + {{{sqrt(3)}}} + . . . + {{{sqrt(n)}}}.


Replace every &nbsp;{{{sqrt(k)}}}&nbsp; by the smaller quantity of &nbsp;1, &nbsp;and you will get this inequality.



Second, &nbsp;1 + {{{sqrt(2)}}} + {{{sqrt(3)}}} + . . . + {{{sqrt(n)}}} <= {{{(n(n+1))/2}}}. 


Replace every &nbsp;{{{sqrt(k)}}}&nbsp; by the greater quantity of &nbsp;<B>k</B>, &nbsp;and you will get this inequality. &nbsp;(You will have the sum of first &nbsp;<B>n</B>&nbsp; natural numbers which is exactly &nbsp;{{{(n(n+1))/2}}}).