Prove that the algebraic expressions are equivalent. 
$$\frac{3^{k+1}-1}{2} + 3^{k+1} = \frac{3^{k+2}-1}{2}$$

with steps make left hand side = right hand side by modifying one or both expressions
Thanks for your help guys, I solved it like this:

$$\frac{3^{k+1}-1}{2} + 3^{k+1} = \frac{3^{k+1}-1}{2} + \frac{2*3^{k+1}}{2} = \frac{3^{k+2}-1}{2} $$


$${3^{k+1}-1\over 2}+3^{k+1}={3^{k+1}\over 2}-{1\over 2}+{2(3^{k+1})\over 2}=({3\over 2})3^{k+1}-{1\over 2}={3^{k+2}-1\over 2}$$
