# Mathematical Induction with summations

• Jun 9th 2013, 10:44 AM
mcolula
Mathematical Induction with summations
Hi, i've been solving some excersices but I am stuck with this one.

Proof by Mathematical Induction the following expresion

Attachment 28556

Well the first thing I did was rewrite the expesion as this:

Attachment 28559

then i proof this is true for p(1)

Attachment 28558

then I assume this true for p(k)

Attachment 28560

p(k+1)

Attachment 28561

then

Attachment 28562

replacing Sk

Attachment 28563

and then i did the sum

Attachment 28564

after this step I don't really know what to do for symplify the expression, so if can help it would be great and if i made a mistake in previous steps please make know. thanks :)
• Jun 9th 2013, 11:04 AM
Plato
Re: Mathematical Induction with summations
Your common denominator is incorrect. It should be just $(k+3)!$
• Jun 9th 2013, 03:55 PM
mcolula
Re: Mathematical Induction with summations
doing (k+3)! the common denominator, the resultant expression is

Attachment 28566

is there a way in which I can simplify this expression to get this result?

Attachment 28567
• Jun 9th 2013, 03:59 PM
Plato
Re: Mathematical Induction with summations
Quote:

Originally Posted by mcolula
doing (k+3)! the common denominator, the resultant expression is
Attachment 28566
is there a way in which I can simplify this expression to get this result?
Attachment 28567

\begin{align*} \frac{1}{2} - \frac{{n + 1}}{{\left( {n + 2} \right)!}} + \frac{{{{(n + 1)}^2} + (n + 1) + 1}}{{(n + 3)!}} &= \frac{1}{2} + \frac{{ - (n + 1)(n + 3) + {{(n + 1)}^2} + (n + 1) + 1}}{{(n + 3)!}}\\ &= \frac{1}{2} - \frac{{(n + 1) + 1}}{{\left[ {(n + 1) + 2} \right]!}}\end{align*}
• Jun 9th 2013, 06:17 PM
mcolula
Re: Mathematical Induction with summations
For those who will be visiting this thread in the future this is the whole answer to the problem

#Problem
Proof by mathematical induction the following expression

Attachment 28568

At first place to do this more easy to understand we rewrite this expression like:

Attachment 28569

and then we will proof this is true for P(1)

Attachment 28570

then we will assume this is true for P(k)

Attachment 28571

and then we will proof thi is true for P(k+1)

Attachment 28572

then we must obtain that Sk+1 is equal to the sum of Sk and Uk+1

Attachment 28573

replacing Sk

Attachment 28574

and doing the sum and keeping the common denominator as (k+3)! we obtain this

Attachment 28575

then we know that

Attachment 28577

the answer continues on the next post
• Jun 9th 2013, 06:36 PM
mcolula
Re: Mathematical Induction with summations
then we obtain this expression

Attachment 28580

and factoring and doing some algebra

Attachment 28581

Attachment 28582

Attachment 28583

Attachment 28584

Attachment 28585

Now we have proven S is true by mathematical induction.

I hope this is useful in the future and if there is some kind of language error make know to correct it thanks.