What!? This Easy!?

Here's how you do it mentally.

$1+2+3+4+5+6+6+8+9+10+11+12$

$=(1+12)+(2+11)+(3+10)+(4+9)+(5+8)+(6+6)$

$=5 \times (13) +12$

$=65+12 = \boxed {77}$

If you don't like that,

$1+2+3+4+5+6+6+8+9+10+11+12$

$= 1+2+3+4+5+6+7+8+9+10+11+12-1$

$= (1+12)+(2+11)+(3+10)+(4+9)+(5+8)+(6+7)-1$

$= 6 \times 13 - 1$

$= 78 - 1 = \boxed {77}$

(13x5)+12=77

Actually the main thing was that it was 6 in place of 7

that will be- $\frac{n(n+1)}{2}-1$ $=\frac{12(12+1)}{2}-1$ $=78-1=77$

LaTex: $\large 1+2+3+4+5+6+6+8+9+10+11+12 = (5*13) + 12 = 77$

That was a simple one.

→1 + 2+3+4+5+6+(6+1)+8+9+10+11+12 - 1

→1+2+..............................+n= n (n+1)/2.

→(1+2+3+...............................+12) - 1 = 12×13/2 - 1= 78 -1=77

First no + last no = 13 and the middle sum is 12; thus 13 comes 5 times and then add 12 to the product

I do it in a lazy and quick fashion :) Try to find number pairs that can be easily added first..eg., (12+8) + (11+9) + (10) + (6+4) + (6+5) + (2+3+1)