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- Thread starter Guilmon
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If you theoretically could line up a row that is infinitely long, then you know that any domino will push the next one over, as long as the first domino is pushed. Mathematical induction works the same way. You need to prove a statement for a base case (which is equivalent to pushing the first domino), and then you need an inductive step, which is to prove that IF the statement is true for an arbitrary case, THEN the statement will be true for the next (which is equivalent to any domino pushing the next one over).Verify that

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

is true using induction

This problem has me stumped...

So how do you think you would go about proving the base case?

- Feb 13, 2012

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With simple steps You can verify that is...Verify that

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

is true using induction

This problem has me stumped...

$(n+1)!-1 = (n+1)\ n! -1 = n\ n! + n! -1$ (1)

What does You suggest (1)?...

Kind regards

$\chi$ $\sigma$

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After having verified $P_1$ is true, I would next state the induction hypothesis $P_k$:

\(\displaystyle \sum_{i=1}^k(i\cdot i!)=(k+1)!-1\)

Now, our goal is to algebraically transform $P_k$ into $P_{k+1}$.

What do you think we should do to both sides of $P_k$ to meet that goal?

\(\displaystyle \sum_{i=1}^k(i\cdot i!)=(k+1)!-1\)

Now, our goal is to algebraically transform $P_k$ into $P_{k+1}$.

What do you think we should do to both sides of $P_k$ to meet that goal?

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- Jan 30, 2012

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I am still not sure that you understand how induction works andThe nth case I understand. It is the n+1th case that has me stumped.

To make sure we are on the same page, I recommend starting from the very beginning: identifying the property P(n) that you need to prove for all n. Without this step, everything else is useless. Note that P(n) has to be true or false for each concrete n; in particular, P(n) cannot be a number such as 1(1!)+2(2!)+...+n(n!). Identifying P(n) in simple cases is easy: just remove the words "for all n" (if they are present) from the claim you are asked to prove. In this case, P(n) is the equality 1(1!)+2(2!)+...+n(n!) = (n+1)! - 1.

Next, you need to know how to write P(0) (or P(1)) and P(n+1). To do this, replace n with 0 (or 1, or n+1, respectively) everywhere in P(n).

Now you should be able to write the base case P(0) (or P(1)) and the induction step: "For all n, P(n) implies P(n+1)". Write both claims explicitly, replacing P by its definition. Only when you do this, you'll know