nineteen

using any four consecutive digits along with any operations and brackets
how can you make 19?

starting with 1 (i.e. using 1,2,3,4): 

• (2 + 4) x 3 + 1
• 4 x 5 – 3 + 2
• 2 + 3 x 4 + 5

and look for other ways, using four consecutive digits (to make 19):
 
starting with 3: 3 x 6 + 5 – 4
starting with 4: 5 x 6 – 4 – 7   or  4 x 5 + 6 – 7
starting with 5: (8 – 6) x 7 + 5
starting with 6: 9 x 8 ÷ 6 + 7  or  8 ÷ 6 x 9 + 7 (a tough one to find)

it is interesting when there are two options (as there are when starting with a 4) 
so 5 x 6 - 4  =  4 x 5 + 6
are there other examples where this works?

yes, for three consecutive digits, starting with 3:
4  x  5 – 3 = 17
and 3  x  4 + 5 = 17

explore a generalisation for such results
and try to prove it
for any three consecutive numbers . . .

this might involve a numerical explanation, viewing 5 x 6 as 4 x 5 + 10 (and then 10 - 4 is 6)

and breaking down 4 x 5 as 3 x 4 + 8 (and then 8 - 3 is 5) in the above two examples

or it could involve algebra, expanding brackets:

(n + 1)(n + 2) - n

n(n + 1) + (n + 2)