If are of the form for some natural number , determine the value of that would make
the smallest possible positive integer.
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(1)Frist notice that this series in an Arithmetic progression so that the sum of the n first terms is (a(1)+a(n))×n/2.
(2) the sum:a(2)+a(3)....a(99)=a(1)+a(2)....+a(100)-(a(1)+a(100)). which gives us the sum of the first 100 terms minos the first and the 100th term: this sum is equal ,according to (1), to: ((a(1)+a(100))*100/2-(a(1)+a(100))=49(a(1)+a(100)). let a(1)+a(100) be X. so pluging it in the given expression: (49X)^0.5-X^0.5=6X pluging in a(1) and a(100) according to their definition by the series a(i)=n+i: 6X=6(n+1+n+100)^0.5=((2n+101)^0.5)×6 we want this to be minimal Integer so (2n+101)^0.5 has to be minimal Integer so we must find the smallest square that can be written as 2n+101 and this square is 11^2=121=101+2×10 which gives us: n=10 a(89)=10+89=99