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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

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

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then i proof this is true for p(1)

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then I assume this true for p(k)

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p(k+1)

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then

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replacing S_{k}

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and then i did the sum

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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 :)

Re: Mathematical Induction with summations

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

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Re: Mathematical Induction with summations

Thanks for your answer I just have one doubt more

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

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

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Re: Mathematical Induction with summations

Quote:

Originally Posted by

**mcolula** Thanks for your answer I just have one doubt more

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?

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$\displaystyle \begin{align*} \frac{1}{2} - \frac{{n + 1}}{{\left( {n + 2} \right)!}} + \frac{{{{(n + 1)}^2} + (n + 1) + 1}}{{(n + 3)!}} &= \frac{1}{2} + \frac{{ - (n + 1)(n + 3) + {{(n + 1)}^2} + (n + 1) + 1}}{{(n + 3)!}}\\ &= \frac{1}{2} - \frac{{(n + 1) + 1}}{{\left[ {(n + 1) + 2} \right]!}}\end{align*}$

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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

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

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

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

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

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then we must obtain that S_{k+1} is equal to the sum of S_{k} and U_{k+1}

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replacing S_{k}

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

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then we know that

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the answer continues on the next post

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Re: Mathematical Induction with summations

then we obtain this expression

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and factoring and doing some algebra

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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.