Evaluate the sum. Evalute the following sum: $ 1 + 2 + 2 + 3 + 4 + 4 + 5 + 6 + 6 + . . . .+ (n-1) + n + n$. 
I tried doing it but I keep getting the wrong answer. I've used known sums to solve it.
 A: Assuming $n$ is even
$$S=1 + 2 + 2 + 3 + 4 + 4 + 5 + 6 + 6 + . . . .+ (n-1) + n + n$$
$$S=(1+2+3+\cdots+(n-1)+n)+(2+4+6+\cdots+n)$$
$$S=\sum_{k=1}^{n} k +\sum_{k=1}^{n/2} 2k$$
$$S=\frac{n(n+1)}{2}+\frac{n}{2}\left(\frac{n}{2}+1\right)=
\frac{2n(n+1)}{4}+\frac{n(n+2)}{4}
=\frac{3n^2+4n}{4}$$
A: We can group, $$\sum_{r=1}^m(2r-1)+2\sum_{r=1}^m(2r)$$ if $n=2m$
Or
$$\sum_{r=1}^m(2r-1)+2\sum_{r=1}^m2(r-1)$$ if $n=2m-1$
A: Reverse the sequence and add term-wise to the original:
$$ 1 + 2 +   2   + 3 + 4 + 4 + 5 + 6 + 6 + . . . .+ (n-1) + n + n$$
$$ n + n + (n-1) + . . . . + 6 + 6 + 5 + 4 + 4 + 3 +  2   + 2 + 1$$ 
All triples will sum to $(n+1)+(n+2)+(n+1)=3n+4$, and each number is counted twice. But, beware, $n$ must be even and there are only $\frac n2$ triples !
$$S_n=\frac12\frac n2(3n+4)=\frac{3n^2+4n}4.$$
Check: $S_2=5$, $S_4=16$, $S_6=33$...
For odd $n$, we have to infer that the sum stops at the odd term, and
$$S_n=S_{n-1}+n=\frac{3(n-1)^2+4(n-1)}4+n=\frac{3n^2+2n-1}{4}.$$
Check: $S_1=1$, $S_3=8$, $S_5=21$...
