Very Simple Triangles

Pyth, 44 42

ItQpdd*\_*4/Q2)jbms<*dQhQ,c" /\ "2,\/"__\\

The first line:

ItQpdd*\_*4/Q2)
ItQ           )    If the input is not 1
   pdd             Print two spaces
      *\_*4/Q2     Then groups of 4 underscores, repeated input/2 times.

The other two lines are generated by noticing that the second line consists of " /" and "\ " alternating input + 1 times, and the third line consists of "/" and "__\" alternated in the same fashion.

SQL, 182 175 173 187 bytes

Not that this'll ever be the shortest, but it's still amusing to try to minimize sql ;) lol I did this in Oracle 11, however, these should be basic SQL. [edit] as pointed out, I didn't apply the when input = 1 rule - only show 2 lines. can't think of a better way to do it, however, I did save a couple bytes by modifying the v logic ;) adding 2 ahead of time saves a couple bytes by not having to repeat it later [/edit]

select decode(&i,1,'',rpad('  ',v,'____')||z)||rpad(' /',v,'\  /')||decode(y,1,'\')||z||rpad('/',v-1,'__\/')||decode(y,1,'__\')from(select 2+floor(&i/2)*4v,mod(&i,2)y,chr(10)z from dual);

[edit1] removed some unnecessary spaces[/edit1] [edit2] changed &&i to just &i. It cuts down 2 chars, but forces user to input the # of triangles twice ... :P I realized my "good coding habits" using &&i were costing be 2 bytes!! The horror!! [/edit2]

Explanation (note: I use &&1 in this explanation so it only prompts once, the &1 above saves code space, but prompts multiple times ;) )

 select  -- line 1
     decode(&&1,1,'',   -- don't need line 1 if input is 1
     rpad('  ',v,'____') || z ) || -- every pair of triangles
     -- line 2
     rpad(' /',v,'\  /') ||  -- every pair of triangles
          decode(y,1,'\') || z || -- add the final triangle, input: 1,3,5 etc.
     -- line 3
     rpad('/',v-1,'__\/') ||  -- every pair of triangles
          decode(y,1,'__\')   -- add the final triangle, input: 1,3,5 etc.
from (select 2+floor(&&i/2)*4 v,   -- common multiplier. 4 extra chars for every triangle pair
             mod(&&i,2) y,  -- Flag for the final triangle (odd inputs, 1,3,5, etc)
             chr(10) z  -- CR, here to save space.
        from dual);

Output

  SQL> accept i
  1
  SQL> /

   /\
  /__\


  SQL> accept i
  2
  SQL> /

    ____
   /\  /
  /__\/


  SQL> accept i
  3
  SQL> /

    ____
   /\  /\
  /__\/__\


  SQL> accept i
  12
  SQL> /

    ________________________
   /\  /\  /\  /\  /\  /\  /
  /__\/__\/__\/__\/__\/__\/


  SQL>

Python 2, 89 88 87 85 83 named / 81 unnamed

f=lambda n:1%n*("  "+n/2*4*"_"+"\n")+(" /\ "*n)[:2+2*n]+"\n"+("/__\\"*n)[:n-~n+n%2]

(Thanks to @orlp for a byte, and @xnor for another three)

This is a function which takes in an int n and returns the triangles as a string using the row-by-row approach.

e.g. print f(10) gives

  ____________________
 /\  /\  /\  /\  /\  /
/__\/__\/__\/__\/__\/

For the first row, instead of (n>1)* we use 1%n*, since 1%n is 0 if n == 1 and 1 if n > 1.