\frac{-(2n+3)}{4 \cdot 3^n} + \frac{n+1}{3^{n+1}}= \frac{-3(2n+3)+4(n+1)}{4 \cdot 3^{n+1}}=\frac{-(2n+5)}{4 \cdot 3^{n+1}}= - \frac{2(n+1)+3}{4 \cdot 3^{n+1}}