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.

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: $S_1=1$, $S_3=8$, $S_5=21$...
