prove by induction: $3 + 5 + 7 + ... + (2n+1) = n(n+2)$ 
Use the principle of mathematical induction to prove that $$3 + 5 + 7 + ... + (2n+1) = n(n+2)$$ for all n in $\mathbb N$.

I have a problem with induction.  If anyone can give me a little insight it would be helpful.  
 A: If you already know that
$$1+2+3+...+n=\frac{n(n+1)}2$$
we can try the following alternative approach:
$$3+5+7+\ldots+(2n+1)=$$
$$=1+2+3+4+5+\ldots+(2n+1)+(2n+2)-1-2(1+2+3+\ldots+ n+n+1)=$$
$$=\frac{(2n+2)(2n+3)}2-1-2\frac{(n+1)(n+2)}2=(n+1)(2n+3)-1-(n+1)(n+2)=$$
$$=(n+1)\left[2n+3-n-2\right]-1=(n+1)^2-1=n^2+2n=n(n+2)$$
A: I can't comment, but here's a little outline. Let $\displaystyle f(n) = \sum_{k=1}^{n}(2k+1).$ Then $f(1) = 3$.
And $\displaystyle f(n+1) = \sum_{k=1}^{n+1}(2k+1) = \sum_{k=1}^{n}(2k+1)+2(n+1)+1 = f(n)+2n+3. $
A: See the answers on this and read this. 
A: What you wrote above for the base case would work, assuming you state something to the effect that for $n = 1$, $2n + 1 = 2(1) + 1 = 3 = 1(1 + 2)$.  That is, you want to get the $2n + 1$ in there.
For the the induction hypothesis, you can proceed like so:  Assume that the identity is true for all $n \leq k$.  In this case, that means
$$3 + 5 + \ldots + (2k + 1) = k(k + 2)$$
Then you want to see if the identity holds for $k + 1$.  So:  What happens now if you add $2k + 3$ to both sides?  Can you factor the new right-hand side in any sort of helpful way?
If it turns out that, whenever the identity holds true for $n=k$, it also holds true for $n = k + 1$, then you can say it's true for all natural numbers.
