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 June 1st, 2012, 06:22 PM #1 Newbie   Joined: Jun 2012 Posts: 16 Thanks: 0 Induction Hi Try to Figure out the next step in this problem, I have the answer but I don't know how to arrive at it 1x2+2x3+3x4+...+n(n+1)=(n(n+1)(n+2))/3 n=1, 1x2=1(1+1)(1+2) 2=6/3 2=2 thus the statement holds for n=1 Assume true for n=k 1x2+2x3+3x4+...+k(k+1)=(k(k+1)(k+2))/3 Prove true for n=k+1 1x2+2x3+3x4+...+k(k+1)+(k+1)(k+2)=((k+1)(k+2)(k+3) )/3 ((k(k+1)(k+2))/3) +(k+1)(k+2)=((k+1)(k+2)(k+3))/3 ((k(k+1)(k+2))/3) +(3(k+1)(k+2)/3)=((k+1)(k+2)(k+3))/3 This is as far as I can go, and I'm not sure if the steps I have taken are correct (I think I forgot something on the right side) can someone please explain how to arrive at ((k+1)(k+2)(k+3))/3=((k+1)(k+2)(k+3))/3 Then more detailed the explanation the better Thankyou
 June 1st, 2012, 07:05 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs Re: Induction You showed the base case is true, so state your induction hypothesis $P_k$: $1\cdot2+2\cdot3+3\cdot4+\cdots+k(k+1)=\frac{k(k+1) (k+2)}{3}$ Add $(k+1)(k+2)$ to both sides: $1\cdot2+2\cdot3+3\cdot4+\cdots+k(k+1)+(k+1)(k+2)=\ frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$ Factor on the right side: $1\cdot2+2\cdot3+3\cdot4+\cdots+k(k+1)+(k+1)(k+2)=( k+1)(k+2)$$\frac{k}{3}+1$$$ $1\cdot2+2\cdot3+3\cdot4+\cdots+k(k+1)+(k+1)(k+2)=( k+1)(k+2)$$\frac{k+3}{3}$$$ $1\cdot2+2\cdot3+3\cdot4+\cdots+k(k+1)+(k+1)(k+2)=\ frac{(k+1)(k+2)(k+3)}{3}$ $1\cdot2+2\cdot3+3\cdot4+\cdots+(k+1)((k+1)+1)=\fra c{(k+1)((k+1)+1)((k+1)+2)}{3}$ We have derived $P_{k+1}$ from $P_k$ thereby completing the proof by induction.
June 1st, 2012, 07:09 PM   #3
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Re: Induction

Quote:
 Originally Posted by Tommy_Gun Hi Try to Figure out the next step in this problem, I have the answer but I don't know how to arrive at it 1x2+2x3+3x4+...+n(n+1)=(n(n+1)(n+2))/3 n=1, 1x2=1(1+1)(1+2) 2=6/3 2=2 thus the statement holds for n=1 Assume true for n=k 1x2+2x3+3x4+...+k(k+1)=(k(k+1)(k+2))/3 Prove true for n=k+1 1x2+2x3+3x4+...+k(k+1)+(k+1)(k+2)=((k+1)(k+2)(k+3) )/3 ((k(k+1)(k+2))/3) +(k+1)(k+2)=((k+1)(k+2)(k+3))/3 ((k(k+1)(k+2))/3) +(3(k+1)(k+2)/3)=((k+1)(k+2)(k+3))/3 This is as far as I can go, and I'm not sure if the steps I have taken are correct (I think I forgot something on the right side) can someone please explain how to arrive at ((k+1)(k+2)(k+3))/3=((k+1)(k+2)(k+3))/3 Then more detailed the explanation the better Thankyou
Alrighty, well once you assume that it's true for $n=k$, you want to show that the sum $1x2+2x3+...+k(k+1)+(k+1)(k+2)$ is equal to the product $\frac{(k+1)(k+2)(k+3)}{3}$. You do that by using the assumption that it's true for $n=k$ so that $[1x2+2x3+...+k(k+1)]+(k+1)(k+2)=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$. Now that you've done that, it's just a bit of short algebra to reach the final goal. We see that $\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)=\frac{k(k+1)(k+2) }{3}+\frac{3(k+1)(k+2)}{3}=\frac{(k+1)(k+2)(k+3)}{ 3}$.

June 1st, 2012, 07:59 PM   #4
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Re: Induction

Quote:
Originally Posted by Xhin
Quote:
 Originally Posted by Tommy_Gun Hi Try to Figure out the next step in this problem, I have the answer but I don't know how to arrive at it 1x2+2x3+3x4+...+n(n+1)=(n(n+1)(n+2))/3 n=1, 1x2=1(1+1)(1+2) 2=6/3 2=2 thus the statement holds for n=1 Assume true for n=k 1x2+2x3+3x4+...+k(k+1)=(k(k+1)(k+2))/3 Prove true for n=k+1 1x2+2x3+3x4+...+k(k+1)+(k+1)(k+2)=((k+1)(k+2)(k+3) )/3 ((k(k+1)(k+2))/3) +(k+1)(k+2)=((k+1)(k+2)(k+3))/3 ((k(k+1)(k+2))/3) +(3(k+1)(k+2)/3)=((k+1)(k+2)(k+3))/3 This is as far as I can go, and I'm not sure if the steps I have taken are correct (I think I forgot something on the right side) can someone please explain how to arrive at ((k+1)(k+2)(k+3))/3=((k+1)(k+2)(k+3))/3 Then more detailed the explanation the better Thankyou
Alrighty, well once you assume that it's true for $n=k$, you want to show that the sum $1x2+2x3+...+k(k+1)+(k+1)(k+2)$ is equal to the product $\frac{(k+1)(k+2)(k+3)}{3}$. You do that by using the assumption that it's true for $n=k$ so that $[1x2+2x3+...+k(k+1)]+(k+1)(k+2)=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$. Now that you've done that, it's just a bit of short algebra to reach the final goal. We see that $\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)=\frac{k(k+1)(k+2) }{3}+\frac{3(k+1)(k+2)}{3}=\frac{(k+1)(k+2)(k+3)}{ 3}$.
Sorry but I'm still confused, I don't understand the algebraic steps in this part $\frac{k(k+1)(k+2)}{3}+\frac{3(k+1)(k+2)}{3}=\frac{ (k+1)(k+2)(k+3)}{3}$

 June 1st, 2012, 08:07 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs Re: Induction It can be shown by factoring very similarly to what I did in my post above: $\frac{k(k+1)(k+2)}{3}+\frac{3(k+1)(k+2)}{3}=\frac{ (k+1)(k+2)}{3}$$k+3$$=\frac{(k+1)(k+2)(k+3)}{3}$
 June 1st, 2012, 08:21 PM #6 Newbie   Joined: Jun 2012 Posts: 16 Thanks: 0 Re: Induction I get IT now!!! A big thank you to both of you!!!
 June 1st, 2012, 08:23 PM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs Re: Induction Glad to help and welcome to the forum!
 June 1st, 2012, 08:48 PM #8 Newbie   Joined: Jun 2012 Posts: 16 Thanks: 0 Re: Induction $1+a+...+a^n=\frac{1-a^(n+1)}{1-a}$ $\frac{k(k+1)(k+2)}{3}+\frac{3(k+1)(k+2)}{3}=\frac{ (k+1)(k+2)}{3}$$k+3$$=\frac{(k+1)(k+2)(k+3)}{3}$
June 1st, 2012, 08:56 PM   #9
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Re: Induction

Quote:
 Originally Posted by Tommy_Gun $1+a+...+a^n=\frac{1-a^(n+1)}{1-a}$ $\frac{k(k+1)(k+2)}{3}+\frac{3(k+1)(k+2)}{3}=\frac{ (k+1)(k+2)}{3}$$k+3$$=\frac{(k+1)(k+2)(k+3)}{3}$
Please ignore that post... I don't know how to edit a post on this forum

$1+a+...+a^n=\frac{1-a^n^+^1}{1-a}$
Prove by induction

the worked example for this exercise is
assume n=k is true
$1+a+...+a^k=\frac{1-a^k^+^1}{1-a}$
then n=k+1 must also be true
$1+a+...+a^k^+^1=\frac{1-a^k^+^1}{1-a}+ a^k^+^1$

why in the last step did he add $a^k^+^1$ to the right hand side
Shouldn't it be:
$1+a+...+a^k+a^k^+^1=\frac{1-a^k^+^1}{1-a}$

 June 1st, 2012, 09:06 PM #10 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs Re: Induction Your last statement is (almost) where we want to wind up. After having demonstrated the base case is true, we state the induction hypothesis $P_k$ $1+a+\cdots+a^k=\frac{1-a^{k+1}}{1-a}$ Now, we must arrive at $P_{k+1}$ algebraically, so we add the same thing to both sides, in this case $a^{k+1}$: $1+a+\cdots+a^k+a^{k+1}=\frac{1-a^{k+1}}{1-a}+a^{k+1}$ $1+a+\cdots+a^k+a^{k+1}=\frac{1-a^{k+1}}{1-a}+\frac{(1-a)a^{k+1}}{1-a}$ $1+a+\cdots+a^k+a^{k+1}=\frac{1-a^{k+1}+(1-a)a^{k+1}}{1-a}$ $1+a+\cdots+a^k+a^{k+1}=\frac{1-a^{k+1}+a^{k+1}-a^{k+2}}{1-a}$ $1+a+\cdots+a^k+a^{k+1}=\frac{1-a^{k+2}}{1-a}$ $1+a+\cdots+a^{k+1}=\frac{1-a^{(k+1)+1}}{1-a}$ We have derived $P_{k+1}$ from $P_k$ thereby completing the proof by induction. You see, with induction, we must demonstrate that we can, using legal algebraic operations, get from $P_k$ to $P_{k+1}$ which means it must be true for any k. By the way, once you have made 20 or more posts, you will be able to edit your posts. It is an unfortunate anti-spam measure we've had to adopt. We apologize for the inconvenience.

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