Mathematical Inductions..

[Idealistic]

Im' fairly decent with these however there's one problem that seemes to be giving me some grief. I have the first two steps but the third is killing me. Here's the problem:

1(1!) + 2(2!) + 3(3!) + ... + n(n!) = (n + 1)! - 1, for n € N

Step One: 1(1!) = (1 + 1)! - 1
1 = 2 - 1
1 = 1
QED

Step Two: k(k!) = (k + 1)! - 1
QED

Step Three: *Now this is the hard part. I'm going to set it up according to how i think it should be before the real math begins. If I'm wrong just let me know.

[(k + 1)! - 1] + (k + 1)(k + 1)! = (k + 2)! - 1

*Just to explain my setup in Step Three, I just subsituted "k(k!)" for the term it equalled in Step Two for my "k" term, then i added it to the next term "k + 1". I did this but simply putting in "k + 1" for every "k" value on the left side of the equation in Step Two. Then I replaced all of the "k's" on the right side of the equation in Step Two with "k +1" and placed it on the right side of the equation in Step Three.

 

[stapel]

Step Two: k(k!) = (k + 1)! - 1
QED

No and no. You aren't trying to prove that k(k!) = (k + 1)! - 1; you're trying to prove that 1(1!) + 2(2!) + ... + k(k!) = (k + 1)! - 1. So your assumption steps needs to be corrected.

Also, you are assuming this part, so you can't say "QED" (thus is proved, etc) for this bit.

[(k + 1)! - 1] + (k + 1)(k + 1)! = (k + 2)! - 1

Start with the formula you're actually working with:

. . . . .1(1!) + 2(2!) + 3(3!) + ... + k(k!) + (k + 1)((k + 1)!

Now substitute the (corrected) assumption. Factor out the (k + 1)! part, and simplify what remains. :wink:

Eliz.

 

[Idealistic]

Hmmmmm. I see, I see. Well, The only thing I'm not sure of now is if I can still factor out the (k + 1)! if the "-1" is part of that term. Can I take (k + 1)! out of [(k + 1)! -1] + (k + 1)(K + 1)! ? Or is the "-1" not even an exclusive part of the (k + 1)! -1? Im just choking on the basic fundamentals lol.

 

[stapel]

The only thing I'm not sure of now is if I can still factor out the (k + 1)! if the "-1" is part of that term. Can I take (k + 1)! out of [(k + 1)! -1] + (k + 1)(K + 1)! ? Or is the "-1" not even an exclusive part of the (k + 1)! -1?

I'm sorry, but I couldn't understand the above.

Please show your steps, from the assumption-substitution step onwards, showing (with algebra) what you did in attempting to follow the posted instructions. Thank you.

Im just choking on the basic fundamentals lol.

Unfortunately, you need those basic fundamentals from algebra and trig, if you are to succeed in calculus. If you are finding yourself frequently lost with respect to this background material, you might want to think about hiring a tutor; this tutor would then be able, during your face-to-face sessions, to re-teach whatever has been forgotten or misunderstood.

Eliz.

 

[Idealistic]

I'll try to be a little more clear. Becuase 1(1!) + 2(2!) + 3(3!) + ... + k(k!) is equal to [(k + 1)! - 1], I subsitute [(k + 1)! - 1] into the third step in place of k(k!). Now I want to factor out the (k + 1)! out of [(k + 1)! *- 1*] + (k + 1)(k + 1)!. However, because the *- 1* is right after the first "(k + 1)!" I was unsure if I was still able to factor it {(k + 1)!} out. Basically, do I have do do something with the negative one or can I simply place it on the outside of the brackets?

JIC, the astrics are there to highlight the negative one not to show multiplication

for instants, I can't no longer factor out 5 out of this example

[(5 - 1) + 5x], but i can out of this example

(5 + 5x) - 1 = 5(1 + x) - 1

So, does the negative one go in or out of the brackets?

 

[Deleted member 4993]

I'll try to be a little more clear. Becuase 1(1!) + 2(2!) + 3(3!) + ... + k(k!) is equal to [(k + 1)! - 1],

No...may be you are not writing what you are thinking.

The proof should be as follows:

I

1(1!) + 2(2!) + 3(3!)...k(k!) = (k+1)! - 1

suppose k=2

Then

1(1!) + 2(2!)

= 1 + 4

= 5

= 6 - 1

= (2+1)! - 1 .............The given equation is true for k = 2

II

Assume the equation is true for k = n

then

1(1!) + 2(2!) +...n(n!) = (n+1)! - 1

III

For k = n+1

1(1!) + 2(2!) +...n(n!) + (n+1)*[(n+1)!]

= (n+1)! - 1 + (n+1)*[(n+1)!]

= [(n+1)!] * {1 + (n+1)} - 1

= [(n+1)!] * {n+2} - 1

= (n+2)! - 1

Thus

the given equation is true for k = n+1 - when we assume that it is true for k = n.....................(1)

We have proven that it is true for k=2 so by (1) it is true for k = 3

We have proven above that it is true for k=3 so by (1) it is true for k = 4

We have proven above that it is true for k=4 so by (1) it is true for k = 5

and ad infinitum....

I subsitute [(k + 1)! - 1] into the third step in place of k(k!). Now I want to factor out the (k + 1)! out of [(k + 1)! *- 1*] + (k + 1)(k + 1)!. However, because the *- 1* is right after the first "(k + 1)!" I was unsure if I was still able to factor it {(k + 1)!} out. Basically, do I have do do something with the negative one or can I simply place it on the outside of the brackets?

JIC, the astrics are there to highlight the negative one not to show multiplication

for instants, I can't no longer factor out 5 out of this example

[(5 - 1) + 5x], but i can out of this example

(5 + 5x) - 1 = 5(1 + x) - 1

So, does the negative one go in or out of the brackets?

 

[stapel]

Becuase 1(1!) + 2(2!) + 3(3!) + ... + k(k!) is equal to [(k + 1)! - 1], I subsitute [(k + 1)! - 1] into the third step in place of k(k!).

No; (k + 1)! - 1 is not equal to k(k!), so it should not be put in place of only that one term. The (k + 1)! - 1 should replace the entire sum from 1(1!) through k(k!); not just the last term in that sum.

Now I want to factor out the (k + 1)! out of [(k + 1)! *- 1*] + (k + 1)(k + 1)!.

Since, in your final form, you need to have a "-1", and since (k + 1)! is not a factor of 1, you probably don't want to try factoring it out of that term. Instead, try factoring (k + 1)! out of the terms that do contain it, and simplifying the result.

. . . . .[(k + 1)! - 1] + (k + 1)(k + 1)!

. . . . .(k + 1)! - 1 + (k + 1)(k + 1)!

. . . . .[(k + 1)! + (k + 1)(k + 1)!] - 1

. . . . .(k + 1)! [1 + (k + 1)] - 1

...and so forth.

Eliz.

 

[Idealistic]

thanx alot for clearing that up. We went over it today in class, and now when I read what you were telling me is a lot clearer now.

My appreciation goes out to all that helped.