Question
a sequence is formed by writting 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 800th term
Answer
<P>It looks like a sequence of one 1 followed by two 2's followed by three 3's and so on followed by n n's.<BR><BR>So what would it be?<BR><BR>The values change at terms 1, 3, 6, 10, 15, 21, ....<BR><BR>Actually this problem was first solved by Karl Friedrich Gauss. According to legend, when he was a child, his teacher wanted to keep the children busy for some time, so he asked them to add up all positive integers between 1 and 100, inclusive. Perhaps your high school math teacher has already told you the story. <BR><BR>Young Karl took only a few minutes to solve it. <BR><BR>He realized that <BR><BR>1 + 2 + 3 + ...... + 100 = X<BR>also that<BR>100 + 99 + 98 + .... + 1 = X<BR><BR>Thus, <BR><BR>101 + 101 + 101 + .... 101 = 2X<BR><BR>There are 100 terms, so,<BR><BR>10100 = 2X<BR>5050 = X<BR><BR>In general (substitute n for 100), <BR><BR>X = 1 + 2 + 3 + ... + n = n(n+1)/2.<BR><BR>You will see that X takes on values where your sequence changes: if n = 1, then X = 1; if n = 2, then X = 3; if n = 3, then X = 6, and so on. <BR><BR>So for what value of n will X equal or exceed 800? <BR><BR>800 = n(n+1)/2. <BR><BR>By inspection, we can see that (39)*(40)/2 = 780 and that (40)*(41)/2 = 820. So the 800th term of the sequence would be 40. <BR><BR><BR>Joe M. <BR>Forest Inventory and Analysis<BR>Knoxville, TN <BR><BR></P>
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