jmc

We have that $$3+\frac{3+k}{4}+\frac{3+2k}{4^2}+\frac{3+3k}{4^3}+\cdots =8.$$Multiplying this equation by 4, we get $$12+(3+k)+\frac{3+2k}{4}+\frac{3+3k}{4^2}+\cdots =32.$$Subtracting these equations, we get $$12+k+\frac{k}{4}+\frac{k}{4^2}+\frac{k}{4^3}+\cdots =24.$$Then $$12+\frac{k}{1-1/4}=24.$$Solving for $k,$ we find $k=\boxed{9}.$