Are you familiar with induction?Quote:

Originally Posted byQuick

If not another excellent thing for you to learn.

(But first you would have to answer the questions I gave you before we can procede to a new topic).

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- Aug 25th 2006, 09:00 AMThePerfectHackerQuote:

Originally Posted by**Quick**

If not another excellent thing for you to learn.

(But first you would have to answer the questions I gave you before we can procede to a new topic). - Sep 6th 2006, 03:18 AMmalaygoela TRY
I followed TPH's advice and see what I got.

Let $\displaystyle S_n=n^2 + n +2$.

I divide the set of n in two parts:

One is of the form 7k-4 and second the rest.

For the first form, you can directly check by substituting the general form 7a-4 for n.

Now the second case.

We could see that $\displaystyle S_1, S_2,S_4,S_5,S_6,S_7$ also satisfy the given condition.

Now,$\displaystyle S_{7+k}=S_k + 7(7+2k+1)$

If you substitute the value of k=1,2,4,5,6,7 you get that $\displaystyle S_8.....S_{14}$ are not divisible by 7.

Continuing this , now you can substitute the value k=8,9,11,12,13,14 and you get the non-divisibility of next six(out of seven, one you have alredy proved) by 7.

In this way you can complete the proof by mathamatical induction.

Bye. - Sep 6th 2006, 02:03 PMQuickWelcome Back
Hello Malay!

I thought I wouldn't hear from you for another 6-9 months? - Sep 8th 2006, 01:27 AMmalaygoelQuote:

Originally Posted by**Quick**

It is fortunate that there is a computer centre in the college where I can aceess you mhf anytime. Whenever I find time from college studies(which I barely get), I try to visit MHF. Although, by mid october, I will have internet in the room.

Keep Smiling

Malay