Mathematical Induction Problem: Finding a Formula for a Sequence of Numbers

[heavysoul93]

Homework Statement

Okay, so I'm going to be completely honest, I am really bad at math, and I have been struggling the past couple of weeks in my Quantitative Reasoning class. I am so lost. I don't know if it's my teacher's teaching method or what, but nothing is clicking for me at the moment, and none of the tutors at my school have even heard of any of this stuff.

I just need help getting this problem started. It says:
Find a formula for the following sequence of numbers:
1,5,12,22,35,51,70,92,117,145,176,210,247,287,330,...

Then it asks to prove the formula is correct by mathematical induction. I think I can do the second part, but I have no clue as to how to come up with a formula.

I appreciate any help. Thanks!

 

[susskind_leon]

I think you're adding 4,7,10,13,16,19,...
So $a_n=a_{n-1}+4+(n-2)*3$ for n>1

 

[Deveno]

I think you're adding 4,7,10,13,16,19,...
So $a_n=a_{n-1}+4+(n-2)*3$ for n>1

i think $$a_n = a_{n-1} + 3(n-1) + 1$$ makes for a nicer recursion formula. (the main problem with your formula is that requires integers for a1, a minor point, but important in programming perhaps).

either way, it appears a logical definition for $a_0$ is 0.

 

[Dickfore]

Try this:
http://oeis.org/search?q=1,5,12,22,...76,210,247,287,330&language=english&go=Search

 

[HallsofIvy]

I think you're adding 4,7,10,13,16,19,...
So $a_n=a_{n-1}+4+(n-2)*3$ for n>1

In fact, the "second difference" (7- 4, 10- 7, 13- 3, 16- 13, 19- 16) is the constant 3 so all succeeding differences are 0. That tells you that $a_n$ is a quadratic function of n. Let $a_n= An^2+ Bn+ C$. Then $a_0= C= 1$, $a_1= A+ B+ 1= 5$, and $4A+ 2B+ 1= 12$ Solve $4A+ 2B= 11$ and $A+ B= 4$ for A and B and see if that formula doesn't give the other numbers in the seqeunce.

 

[heavysoul93]

Thanks for the help guys. I'll see what I can do now with the rest of the problem. If I get stuck I'll post back here. Thanks again.