Draw some mountain peaks

Charcoal, 16 bytes

ＮλＦλ«Ｐ↘⁻λι←↙¹‖Ｔ→

Try it online!

How?

Ｎλ inputs the size of the largest mountain into λ. Ｆλ« runs a loop over values of ι from 0 through λ-1. (The closing » is implied at the end of the program.)

Inside the loop, Ｐ↘⁻λι calculates λ-ι and draws, without moving the cursor afterward, a line of that length going southeast. Based on its direction, this line will consist of \ characters. ← moves one step to the west, and ↙¹ draws a line of length 1 going southwest (made of /). Finally, ‖Ｔ→ horizontally reflects the drawing, transforming characters as appropriate: \ becomes / and / becomes \.

Adding the dump instruction Ｄ at the beginning of the loop (try it) allows us to see the progression:

    /\
   /  
  /   
 /    
/     
    /\    
   /\ \   
  /    \  
 /      \ 
/        \
    /\    
   / /\   
  / /\ \  
 / /    \ 
/ /      \
    /\    
   /\ \   
  / /\ \  
 / /\ \ \ 
/ /    \ \
    /\    
   / /\   
  / /\ \  
 / / /\ \ 
/ / /\ \ \

JavaScript (ES6), 75 bytes

for(n=prompt(s="/\\");n--;s=n%2?s+' \\':'/ '+s)console.log(" ".repeat(n)+s)

The full program is currently slightly shorter than the recursive function:

f=n=>n?" ".repeat(--n)+`/\\
`+f(n).replace(/\S.+/g,x=>n%2?x+" \\":"/ "+x):""

Python 2, 67 bytes

n=input()
s='/\\'
while n:n-=1;print' '*n+s;s=['/ '+s,s+' \\'][n%2]

Prints line by line, accumulating the string s by alternately adding a slash to the left or right based on the current parity of n. Prefixes with n spaces.

An alternative way to update was the same length:

s=n%2*'/ '+s+~n%2*' \\'
s=['/ '+s,s+' \\'][n%2]

A recursive method was longer (70 bytes).

f=lambda n,s='/\\':n*'_'and' '*~-n+s+'\n'+f(n-1,[s+' \\','/ '+s][n%2])