<PRE><B>Question 1184590</B>
&lt;br&gt;
Technically the problem is defective; ANY subsequent numbers would form a valid sequence....&lt;br&gt;
However, there is a clear pattern in the given numbers that makes it possible to find what is almost certainly the intended answer.&lt;br&gt;
Look at the sequence of terms and the sequences of first and second differences:&lt;br&gt;&lt;pre&gt;
   1   7  17  31  49  71
     6  10  14  18  22
       4   4   4   4&lt;/pre&gt;
The constant row of second differences means the sequence can be produced by a second degree polynomial function.&lt;br&gt;
Since we need the 100th term, we need to find the quadratic function.  (Alternatively, we could continue the array of numbers shown, repeating the common difference of 4 as many times as we need to reach the 100th term of the sequence.  But that is not an efficient way to reach the answer!)&lt;br&gt;
One way to find the quadratic function is to use the first three terms to get a system of 3 equations in 3 unknowns:&lt;br&gt;
t(n)=an^2+bn+c&lt;br&gt;
t(1): a+b+c=1
t(2): 4a+2b+c=7
t(3): 9a+3b+c=17&lt;br&gt;
I&#39;ll let you finish the task of finding the quadratic function by that method.&lt;br&gt;
NOTE: It&#39;s a good exercise in formal algebra; I strongly recommend you do it....&lt;br&gt;
I will finish finding the quadratic function by a different method that you might find useful to know.&lt;br&gt;
FACT: The common second difference of 4 means the quadratic function has leading coefficient 4/(2!) = 4/2 = 2.  So the function is&lt;br&gt;
t(n)=2n^2+bn+c&lt;br&gt;
The difference between t(n) and 2n^2 will be a linear function which can easily be determined.&lt;br&gt;&lt;pre&gt;
   n  t(n)  2n^2    2n^2-t(n)
  ----------------------------
   1    1     2      -1
   2    7     8      -1
   3   17    18      -1&lt;/pre&gt;
The difference between t(n) and 2n^2 is the constant -1, so the quadratic function is&lt;br&gt;
t(n)=2n^2-1&lt;br&gt;
You can verify that by using the formula to find the given 4th, 5th, and 6th terms.&lt;br&gt;
ANSWER: the 100th term of the sequence is 2(100^2)-1 = 2(10000)-1 = 19999&lt;br&gt;
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In response to the student&#39;s question....&lt;br&gt;
We can use a, or t, or p, or W, or whatever we want to name terms of the sequence.&lt;br&gt;
I used t(n) because it represents the n-th term (&quot;t&quot; for term).&lt;br&gt;
I specifically did not use a, because I was using a as the leading coefficient of the general quadratic function ax^2+bx+c.&lt;br&gt;
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