Thread:

1. Induction 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

2. Originally Posted by monkbear
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

3. Originally Posted by monkbear
How would I go on to prove this with induction exactly? Thanks
Induction works in two steps. First you prove it true for some base case. In this case, if P(n) reads

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

show that P(0) is true. That should be straight-forward computation. The inductive step is to assume it is true for some $k \geq 0$. Assuming P(k), demonstrate P(k+1). By establishing those two facts regarding P(n), you can prove it is true for every $n\geq 0$. 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?

Originally Posted by Plato
That link didn't work for me Plato (using FF2). Broken?

4. Originally Posted by bryangoodrich
that link didn't work for me plato (using ff2). Broken?
fixed.

5. I've never used Wolfram Alpha before, but I have to say ... that's awesome!

6. Originally Posted by monkbear
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
You should have a closer look at the pattern.

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

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

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

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

To prove using induction that

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

P(k)

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

P(k+1)

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

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

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

which compares to

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

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

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

It remains to test the base case for n = 1, which is clearly true.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9254982471466064, "perplexity_flag": "middle"}