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Problem Presentation

Problem Presentation. Meng Li 09-21-2012. Problem set 2. Solution 1:. Since the problem is to find a closed form representation of the sequence, that is we need to find the relation between “ a n ” and “ n ”.

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Problem Presentation

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  1. Problem Presentation Meng Li 09-21-2012

  2. Problem set 2

  3. Solution 1: • Since the problem is to find a closed form representation of the sequence, that is we need to find the relation between “an” and “n”. • First of all, we need to figure out the domain of the relation, which is the number that n could be. Since a1=1,when n=1; and an=an-1+n, when n>1.So we need to find the an, when n is equal or greater than1. • Put the values of n in the sequence we can get the list as follow:

  4. Solution 1: • a1=1 • a2=a1+2 • a3=a2+3 • … • an-1=an-2+(n-1) • an=an-1+n • Now add all of them, it is clear that all of the red items can be canceled, then we can get: • an = 1+2+3+…+(n-1)+n

  5. Solution 1: • Now we have a new problem, how can we find the sum of 1+2+3+…+(n-1)+n ? Since an=1+2+…+(n-1)+n we can rewrite that: an=n+ (n-1)+ …+ 2 + 1 (each pair of items equal to n+1) an=1+ 2 +… +(n-1)+ n there are n items in total Now we add the two functions together We get:2 an=(n+1) * n So an=1/2 * (n2 + n)

  6. Solution 2 • Assume the relation between an and n is a function: an=f(n)=An2+Bn+C (A,B,C are real numbers) • Then we can get: a1=1 (1,1) a2=1+2=3 (2,3) a3=3+3=6(3,6) Then put them in the function f(n) A+B+C=1 A=1/2 4A+2B+C=3 B=1/2 9A+3B+C=6 C=0 f(n)=1/2 *(n2)+1/2*n So an=1/2*(n2+n)

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