# mathematic induction

• May 8th 2011, 05:35 AM
yrlim1
mathematic induction
hi guys! i need help in solving this induction problem!!
http://i580.photobucket.com/albums/s..._James/010.jpg

I try solving it! but i don't know if is correct very confusing question!

My solution
http://i580.photobucket.com/albums/s..._James/013.jpg

http://i580.photobucket.com/albums/s..._James/014.jpg

http://i580.photobucket.com/albums/s..._James/015.jpg
• May 8th 2011, 08:19 AM

The sum from i=1 to i=2 is 1/3+1/7=10/21
which is <1 so the base case is ok.

You have the P(k) proposition as

$\displaystyle \sum_{i=1}^k\left(\frac{1}{i^2+i+1}\right)\le\ 2-\frac{1}{k-1}$

$\displaystyle \sum_{i=1}^{k+1}\left(\frac{1}{i^2+i+1}\right)\le\ 2-\frac{1}{k+1-1}$

Then you try to prove that IF P(k) is true THEN P(k+1) will be true.

Do this by using the fact that if Pk) is true,
P(k+1) LHS will be less than or equal to

$\displaystyle 2-\frac{1}{k-1}+\frac{1}{(k+1)^2+(k+1)+1}$

and hopefully this will be $\displaystyle \le\ 2-\frac{1}{k}$

$\displaystyle \frac{1}{k^2+2k+1+k+2}\le\ \frac{1}{k-1}-\frac{1}{k}\;\;?$

$\displaystyle \frac{1}{k^2+3k+3}\le\frac{k-(k-1)}{(k-1)k}\;\;?$

$\displaystyle \frac{1}{k^2+3k+3}\le\frac{1}{k^2-k}\;\;?$

$\displaystyle k^2+3k+3\ge\ k^2-k\;\;?$

$\displaystyle 4k+3\ge\ 0\;\;?$
• May 8th 2011, 03:50 PM
yrlim1
hi, so the entire ans i did is wrong?? sorry I really not very good in this! can u show me the correct solution?? so I can see my mistake!!
• May 8th 2011, 04:02 PM
Didn't you add the part in yellow (at the top of your third picture) to both sides in your effort?
If the Left is less than or equal to the Right in that case initially,
it will be afterwards, since you only added the same thing to both sides,
so it's pointless.

You must understand what you are trying to do with the "k" and "k+1" concept.

I've shown you most of the steps.
Study it and try to answer the final line
• May 10th 2011, 04:18 PM
Actually your work is fine (apart from your base case of course)