How do I work out (2+y)^4 using the binomial expansion?

We're going to use the binomial expansion formula, which is a lot easier to remember than it looks:
(a+b)ⁿ=C₀*aⁿ*b⁰ + C₁*a^(n-1)*b¹ + C₂*a^(n-2)*b²+....+ C_n*a⁰bⁿ Now let's apply this formula to the question asked. First, we note that a=2, b=y and n=4. Second, we want to work out the coefficients (C₀,C₁ etc.) and to do this we write out the first 5 rows of Pascal's triangle. This is done by starting with a 1 and then each row below is made up by adding the 2 nearest numbers above it together, (if there is a blank space use 0). So this would be:        
        1
      1   1
    1   2   1
  1   3   3  1
1  4   6   4  1
Where the bottom row are the coefficients for the binomial expansion formula C₀=1,C₁=4,C₂=6,C₃=4,C₄=1
Now all we need to do is apply the binomial expansion formula from above:
(2+y)⁴=12⁴y⁰ + 42³y¹ + 62²y² + 42¹y³ + 12⁰*y⁴
Now simplify and remember that anything to the power 0 is 1, (y⁰=1).
(2+y)⁴=16 + 32y + 24y² + 8y³ + y⁴
and that's it.
NOTE: Instead of trying to remember the complicated formula you may find it easier to notice that for each successive term you subtract 1 from the power of a and add 1 to the power of b.