# Thread: Mathematical Induction with summations

1. ## 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

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

then i proof this is true for p(1)

then I assume this true for p(k)

p(k+1)

then

replacing Sk

and then i did the sum

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

2. ## Re: Mathematical Induction with summations

Your common denominator is incorrect. It should be just $\displaystyle (k+3)!$

3. ## Re: Mathematical Induction with summations

doing (k+3)! the common denominator, the resultant expression is

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

4. ## Re: Mathematical Induction with summations

Originally Posted by mcolula
doing (k+3)! the common denominator, the resultant expression is

is there a way in which I can simplify this expression to get this result?
\displaystyle \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*}

5. ## 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

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

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

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

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

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

replacing Sk

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

then we know that

the answer continues on the next post

6. ## Re: Mathematical Induction with summations

then we obtain this expression

and factoring and doing some algebra

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.