Grades 11-12 Video Solutions 2013-Levels 11&12 Video Solutions 2013 problem29

Video Transcription

Question number 29.  Julian has written an algorithm in order to create a sequence of numbers as a sub 1 is 1, a sub m plus n is given by the formula a sub m plus a sub n plus this product m times n  where m and n are natural numbers. Find the value of the 100th element of the sequence a sub 100.  What we can do is work with the specific choice of n  as n equal to 1 and obtain a simpler recursive formula.  So we set n is equal to 1 and then a sub m plus 1 is equal to a sub m plus a sub n plus m times 1 and that simplifies as a sub m plus 1 plus m.  So we have here the same index on both sides  which gives us a recursive formula for a sub 100  and that would be a 99 plus 100 and then we can keep going,  a 99 would be a 98 plus 99 and then we still have that 100, a 98 would be a sub 97 plus 98 and we still have the 99 and the 100 and we keep going until we exhaust our indices  and we end up with 1 plus 2 plus 3 all the way up to 98 plus 99 plus 100 which is the sum of the first 100 positive integers and that's given by the formula n times n plus 1 over 2.  In our case with n equal to 100, we obtain 5050  and we're almost ready to call this our solution.  I thought circle it as it is the solution but we have to check here that our formula is indeed valid  and we haven't done that yet so we need to check,  we need to set am equal to m times m plus 1 over 2 and check that with that formulation am plus n is indeed equal to a sub m plus a sub n, sorry that should have been an n,  plus the product of m and n and the algebra is available  in the suggested solutions. It's a straightforward but lengthy calculation  which I will not repeat here and just say that the answer is indeed what we think it is, it is e.

Video Summary

Julian's algorithm generates a sequence using a recursive formula. By setting \( n = 1 \), the recursive formula becomes \( a_{m+1} = a_m + m + 1 \). To find the 100th element, \( a_{100} \), you calculate the sum of the first 100 positive integers. Using the formula \( \frac{n(n+1)}{2} \), where \( n = 100 \), the sum is 5050. The process involves verifying that this formulation satisfies the original expression \( a_{m+n} = a_m + a_n + mn \), which it does, confirming the solution is correct. The answer is 5050.

Keywords

Julian's algorithm

recursive formula

sequence

sum of integers

5050