Proof by induction that $(n+1)+(n+2)\cdots+2n=\frac{1}{2}n(3n+1)$ 
Prove by induction that $(n+1)+(n+2)\cdots+2n=\frac{1}{2}n(3n+1)$

I was not really sure how to do this, but I assumed that the case holds for $n=k$, therefore $\displaystyle\sum_{r=1}^kk+r=\frac{1}{2}k(3k+1)$.
Want it on this form:$$\frac{1}{2}(2k+1)(6k+3)=\frac{1}{2}12k^2+12k+3$$
Process:
$$\displaystyle\sum_{r=1}^k(k+r)+(k+k+1)=\frac{1}{2}(3k^2+k)+(2k+1)$$
$$=\frac{3k^2+5k+2}{2}$$
Im very confused here, and I'm sure there are loads of mistakes here but I just can't spot them. Could anyone be so kind to help?
Thanks,
 A: Hint:its easy to prove by induction :$$1+2+3+...+n=\frac{n(n+1)}{2}$$$$(n+1)+(n+2)\cdots+2n=(n+n+...+n)+(1+2+3+..+n)=n^2+\frac{n(n+1)}{2}$$ 
A: Hint: it's easy to verify the result for $n=1$.
Now 
$$((n+1)+1)+((n+1)+2)+\cdots+2(n+1)=(n+2)+(n+3)+\cdots+(2n+2)\\
=\frac{1}{2}n(3n+1)-(n+1)+(2n+1)+(2n+2)$$
and verify that's equal to $\frac{1}{2}(n+1)(3(n+1)+1)$.
A: Your induction hypothesis
$$\sum_{r=1}^k(k+r)=\frac12k(3k+1)$$
is fine, but you went astray at the start of the induction step. The $k+1$ case is 
$$\sum_{r=1}^{k+1}\big((k+1)+r\big)=\frac12(k+1)\big(3(k+1)+1\big)\;,$$
so you should be starting with
$$\begin{align*}
\sum_{r=1}^{k+1}\big((k+1)+r\big)&=\sum_{r=1}^k\big((k+1)+r\big)+\big((k+1)+(k+1)\big)\\
&=\sum_{r=1}^k\big((k+r)+1\big)+2(k+1)\\
&=\sum_{r=1}^k(k+r)+\sum_{r=1}^k1+2(k+1)\\
&=\sum_{r=1}^k(k+r)+k+2(k+1)\\
&=\sum_{r=1}^k(k+r)+3k+2\;.
\end{align*}$$
Now you can apply your induction hypothesis and simplify.
A: Hint:
For the case $n+1$ develop the right part: $1/2(n+1)(3(n+1)+1)$ to come back to the case $n$ plus something and check that whats left is equal to what is in more on the left part for $n+1$ (be careful that there is also one term in less in the sum ...)
A: Now assume that the identity holds true for $n = k$. This implies
$$ \sum_{r=1}^k (k+r) = \frac 12 k(3k+1)$$
Then for $k+1$ we have,
$$\sum_{r=1}^{k+1} (k+1+r) = \sum_{r=1}^{k} (k+1+r) + 2(k+1) = \sum_{r=1}^{k} (k+r) + \sum_{r=1}^k 1 + 2(k+1) $$ $$= \sum_{r=1}^{k} (k+r) + k + 2(k+1) =  \frac 12 k(3k+1) + k + 2(k+1)$$ $$ = \frac 12 (k+1)(3(k+1)+1) $$
A: Check for $n=1$:
$$(n+1)+\cdots+2n=2=\frac12 1(3+1)=2$$
Ok, that holds. Now see that you can prove the case $n+1$ from the case for $n$
$$((n+1)+1)+\cdots +2(n+1)=(n+1)+\cdots+2n - (n+1)+(2n+1)+(2n+2)$$
Here you can use the fact that the formula olds for $n$:
\begin{align}
\ldots &=\frac12 n(3n+1) - (n+1)+(2n+1)+(2n+2)\\
&=\frac12 n(3n+1) + (3n+2)\\
&=\frac12 (3n^2+7n+4)\\
&=\frac12 (n+1)(3(n+1)+1)\\
\end{align}
