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$1+3+5+...+(2n+3)=n^2+4n$

For this series using induction proof.

Base case 1,2,3,.. not working. But induction step works well.

Base case is not given in question.

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closed as unclear what you're asking by D.W., David Richerby, vonbrand, Luke Mathieson, frafl Feb 17 '14 at 9:15

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ What is the question? $\endgroup$ – Dave Clarke Feb 13 '14 at 5:14
  • $\begingroup$ @DaveClarke Base case? or the question incorrect? $\endgroup$ – user5507 Feb 13 '14 at 5:15
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    $\begingroup$ My point is: what is the question? There is no question. Adding a question mark to the title does not make this a question. Make a little effort to write your question properly. $\endgroup$ – Dave Clarke Feb 13 '14 at 5:42
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    $\begingroup$ It is entirely possible for base cases to hold but induction to fail, just as it is possible for base cases to fail but induction to work. In either case, the claim isn't true. Let's prove that $n > n + 1$. Assume that $k > k + 1$ for $k < n$. For $k = n$, we have $n = (n - 1) + 1 > ((n - 1) + 1) + 1 = n + 1$. The inductive step holds, but the claim is false, since there are no base cases that work. $\endgroup$ – Patrick87 Feb 13 '14 at 5:53
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    $\begingroup$ In this case, the correct value for the summation is $1 + 3(n+1) + n(n+1) = 1 + 3n + 3 + n^2 + n = 4 + 4n + n^2 = (n+2)^2$. This appears to work for all values of $n$. Also note that it's the same as what your equation says, plus four. Make of that what you will. $\endgroup$ – Patrick87 Feb 13 '14 at 5:59
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First of all, the right hand side should be $n^2+4n+4$

You can find the generalized term for the left hand side is $2n-1$. Therefore,

$1+3+5+...+(2n-1)+(2n+1)+(2n+3)=n^2+4n+4$, which gives you:

$1+3+5+...+(2n-1)=n^2$.

Then you can apply the mathematical induction to prove it holds for every positive integer $n$.

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  • $\begingroup$ The general term is given as $2n+3$, not $2n-1$, and I don't see what you gain by subtracting $4n+4$ from each side of the equation: it doesn't make anything easier to prove. And, while I grant that the question is unclear, answering a question about how you prove something by induction with, "... then you apply mathematical induction to prove it" just doesn't help. If the asker knew how to prove things by induction, they wouldn't have asked the question in the first place! $\endgroup$ – David Richerby Feb 13 '14 at 10:35
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    $\begingroup$ Thanks! I have to admit the answer is not strict enough. Base on the title, I thought the asker knew the mathematical induction, but was confused why n=1, 2 and 3 do not hold. Therefore I mentioned that if we want to start with n=1, we have to let the generalized term to be $2n-1$, which supports n=1, 2 and 3. I will try to provide a clear answer next time. Thank you for reminding. $\endgroup$ – yuanshang Feb 14 '14 at 8:01
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A proof by induction requires that the base case holds and that the induction step works. If either doesn't work, then the proof is not valid.

It can definitely happen that the induction step works, but not the base case. If that never happened, we'd define induction without the base case.

Example: consider the property “for any integer $n$, in every heap of $n$ pebbles, there are $n+1$ pebbles”.

Proof of the induction step: Suppose that for some integer $n$, in every heap of $n$ pebbles, there are $n+1$ pebbles. The induction step requires us to prove that in every heap of $n+1$ pebbles, there are $(n+1)+1$ pebbles. To prove this, consider any heap of $n+1$ pebbles. Take one pebble from the heap; the heap now contains $n$ pebbles. By the induction hypothesis, that heap contains $n+1$ pebble. Put back the pebble that was taken off; the heap now contains $n+2$ pebbles, Q.E.D.

Of course the base case fails: in a heap of $0$ pebbles, there is not one pebble.

In the case of the summation $1+3+5+\ldots+(2n+3) \stackrel{?}{=} n^2 + 4n$, the inductive case consists of adding $2n+5$ on the left-hand side and $((n+1)^2 + 4(n+1)) - (n^2 + 4n) = 2n+4$ on the right-hand side. This in fact shows that the difference between the left-hand series and the right-hand series is a constant. To show that the two series are equal, you would need to prove that this constant is zero; this is what the base case would establish. However the base case is $1 + 3 \stackrel{?}{=} 0$; this shows that the difference between the left-hand series and the right-hand series is $4$. This gives you the correct formula: $1 + 3 + \ldots + (2n+3) = n^2 + 4n + 4$ (or equivalently $1 + 3 + \ldots + (2n+3) = (n+1)^2$), which you can prove by induction.

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