Solve the following recurrence relations using substitution method

b$)$

i$),f(n)=f(n-1)+n$

ii$),x(n)=3x(n-1)$

for $n>0,f(0)=0$

$x(n)=x(\frac{n}{2})+n$

for $n>1,x(1)=4$

for $n>1,x(1)=1,n=2^k$a$)$