‘A sequence is formed by writing the integers the corresponding number of times as follows : 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, … What is the 800 th term in this sequence?’
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1Have you heard about the triangular numbers? – Arthur Aug 09 '15 at 08:02
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Yea I found that the "n" values are 1, 3, 6, 10, and 15 and these are the triangular numbers. – MathLOL Aug 09 '15 at 08:10
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This is the same as asking for (the index of) the smallest triangular number that is no less than 800 (a triangular number is a positive integer of the form $\frac{n(n+1)}{2}$).
Solving $\frac{x(x+1)}{2} = 800$ gives $x=39.50312, x=-40.50312$.
Since $\frac{39 \times 40}{2} = 780$ and $\frac{40 \times 41}{2} = 820$ we see that the 800th term of the sequence must be 40.
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a(n(n-1)/2 + 1 ) , ... , a(n(n+1)/2) are equal to n
put n=40 : a(781) , ... , a(820) are equal to 40
so a(800) = 40
mahdokht
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