Check My Find Convolution of n+1, 0<=n<=2

[angel23]

find convolution of.


n+1 0<=n<=2
x[n]=

0 otherwise

h[n]= a^n u[n]

solution:
y[n]= a^n + 2a ^n-1 + 3 a^n-2

is it right?

 

[vsage]

Looks to be, except for the part where you seem to ignore what the step function u[n] does to the solution.