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I've gotten to the final step and believe my problem lies within my algebra.

Prove the following:

$1 \times 3 + 2 \times 4 + 3 \times 5 + ... + N(N+2) = \frac{N(N+1)(2N+7)}6$

Here is my show that it is true answer:

$\displaystyle\frac{(K+1)[(k+1)+1][(2(k+1)+7]}6$

Here is where I am stuck:

K(K+1)(2K+7) + 6 (k+1) [(k+1)+2]/6

So all that needs to be done is algebra and moving around the above to show that it is equal to the show that it is true answer above. I just cannot figure it out. I am pretty sure I have the right equation.

Martin Sleziak
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Ryan
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  • Please use the math mode, to write math. For some basic information about writing math at this site see e.g. here, here, here and here. – frabala Mar 03 '14 at 16:54
  • In a proof by induction you must prove two facts : (i) that your property or formula holds for $n=0$ (called the basis of the induction), and (ii) that, assuming that it holds for $n$, then it holds for $n+1$ (called the induction step). So with your problem you must start showing that the formula N(N+2) = N(N+1)(2N+7)/6 holds when $N=0$ (and this is easy; if you think there is a trick, check also with $N=1$), and then go on with the second step. – Mauro ALLEGRANZA Mar 03 '14 at 16:59

2 Answers2

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Let $\displaystyle f(n):1\cdot3+2\cdot4+3\cdot5+\cdots+n(n+2)=\frac{n(n+1)(2n+7)}6$

holds true for $n=m$

$\displaystyle\implies f(m):1\cdot3+2\cdot4+3\cdot5+\cdots+m(m+2)=\frac{m(m+1)(2m+7)}6$

$\displaystyle\implies f(m+1):1\cdot3+2\cdot4+3\cdot5+\cdots+(m+1)(m+1+2)$

$\displaystyle=\frac{m(m+1)(2m+7)}6+(m+1)(m+1+2)$

$\displaystyle=\frac{m+1}6\left[m(2m+7)+6(m+3)\right]$

$\displaystyle=\frac{m+1}6\left[2m^2+13m+18\right]$

Now, $\displaystyle2m^2+13m+18=(2m+9)(m+2)=[2(m+1)+7][(m+1)+1]$

Finally establish the base case i.e., for $n=1$

lab bhattacharjee
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  • just to add that the final term could be easier to see if you use $$(2m+9)(m+2) = \left(2(m+1) + 7\right)\left((m+1) + 1\right)$$ since then the OP can see that it is exactly the same as before irrespective of the shift of $m\rightarrow m+1$ IMO :). – Chinny84 Mar 03 '14 at 17:01
  • The above proof is a special case of telescopic induction. The proof becomes trivial when reformulated this way -- see my answer. – Bill Dubuque Mar 03 '14 at 17:15
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Using the smple method of telescopy, the inductive proof is equivalent to checking the base case, and checking that $\,g(n)-g(n\!-\!1) = n(n+2)\,$ for $\,g(n) = n(n+1)(2n+7)/6.\,$

Remark $\ $ This is what is effectively done in most any proof of this result (e.g. Lab's answer). The advantage of using the telescopic induction framework is that the path to the induction has already been laid out. You don't need to pull your hair out searching for some trick to make the inductive step work. Instead, all you need to do is calculate the difference of two polynomials and verify a polynomial equality - simple high-school algebra. One could not ask for a more simple algorithm.

Learn to master telescopic induction early in your studies, on simple examples like this, since it will be much more difficult to do so later in more complicated contexts which further obfuscate the simple telescopic cancellation which enable this ubiquitous inductive method.

Community
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Bill Dubuque
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  • Beautiful insightful answer! Your posts on telescopic induction are quite inspiring, and I was very happy to apply the technique myself for the first time today. Thank you! – Anant Mar 18 '14 at 16:58
  • @Anant Thanks for your kind words, and thanks for spreading the telescopy gospel. It deserves to be much more widely known. – Bill Dubuque Mar 18 '14 at 17:16