solve
i)an+1-2an=5, n>=0, a0=1
ii)an+1=an+(2n+3), n>=0, a0=1
For the second one, write: $\displaystyle a_n=a_0+\sum_{k=0}^{n-1}(a_{k+1}-a_k)$ (this is a telescopic sum), from which you deduce, using your equation, $\displaystyle a_n=1+\sum_{k=0}^{n-1}(2k+3) = 1 + (n-1)n + 3n$ (remember $\displaystyle \sum_{k=0}^n k=\frac{n(n+1)}{2}$).
For the first one, you know $\displaystyle a_{n+1}-2a_n$, so the telescopic trick doesn't work directly. However, it works for $\displaystyle {a_n\over 2^n}$: we have $\displaystyle \frac{a_{n+1}}{2^{n+1}}-\frac{a_n}{2^n}=\frac{5}{2^{n+1}}$. A telescopic summation gives: $\displaystyle \frac{a_n}{2^n}=\frac{a_0}{2^0}+\sum_{k=0}^{n-1}\left(\frac{a_{k+1}}{2^{k+1}}-\frac{a_k}{2^k}\right)=1+\sum_{k=0}^{n-1}\frac{5}{2^{k+1}}$$\displaystyle =1+\frac{5}{2}\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}$ (remember the sum of terms in a geometric sequence)
Laurent.