Hi I need help in regards with proving this problem.

If the sequence is 1,13,37,73,121

then it should follow like this 1+12+24+36......12n+(12n+12)

For the formula 6n^2 - 6n + 1.

How would I go on to prove this with induction exactly? Thanks

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- May 26th 2011, 12:55 PMmonkbearInduction help
Hi I need help in regards with proving this problem.

If the sequence is 1,13,37,73,121

then it should follow like this 1+12+24+36......12n+(12n+12)

For the formula 6n^2 - 6n + 1.

How would I go on to prove this with induction exactly? Thanks - May 26th 2011, 01:08 PMPlato
Have a look at this Page

- May 26th 2011, 02:08 PMbryangoodrich
Induction works in two steps. First you prove it true for some base case. In this case, if P(n) reads

$\displaystyle 1 + 12 + 24 + ... + 12n + (12n + 12) = 6n^2 - 6n + 1$

show that P(0) is true. That should be straight-forward computation. The inductive step is to assume it is true for some $\displaystyle k \geq 0$. Assuming P(k), demonstrate P(k+1). By establishing those two facts regarding P(n), you can prove it is true for every $\displaystyle n\geq 0$. Do you understand why this two-step process works? In other words, do you know why the principle of mathematical induction (PMI) is a valid rule of inference?

That link didn't work for me Plato (using FF2). Broken? - May 26th 2011, 02:18 PMPlato
- May 26th 2011, 03:14 PMbryangoodrich
I've never used Wolfram Alpha before, but I have to say ... that's awesome!

- May 27th 2011, 04:20 AMArchie Meade
You should have a closer look at the pattern.

$\displaystyle u_1=1=1+12(0)$

$\displaystyle u_2=13=1+12(1)$

$\displaystyle u_3=37=1+12(1)+12(2)$

$\displaystyle u_n=1+12(1)+12(2)+....+12(n-1)$

To prove using induction that

$\displaystyle u_n=1+12(1)+12(2)+....+12(n-1)=6n^2-6n+1$

**P(k)**

$\displaystyle 1+12(1)+12(2)+....+12(k-1)=6k^2-6k+1$

**P(k+1)**

$\displaystyle 1+12(1)+12(2)+.....+12(k-1)+12k=6(k+1)^2-6(k+1)+1\;\;\;?$

Try to show that**IF**P(k) is true**THEN**P(k+1) will also be true.

**Proof**

If P(k) is true, then P(k+1) is

$\displaystyle 1+12(1)+12(2)+...+12(k-1)+12k=6k^2-6k+1+12k=6k^2+6k+1$

which compares to

$\displaystyle 6(k+1)^2-6(k+1)+1=6\left(k^2+2k+1\right)-6k-6+1$

$\displaystyle =6k^2+12k+6-6k-6+1$

$\displaystyle =6k^2+6k+1$

It remains to test the base case for n = 1, which is clearly true.