How would you solve this recurrence equation: $a_{n+1}-2a_{n}=6\cdot 5^n$ for $n\geq 1$ How would you solve $a_{n+1}-2a_{n}=6\cdot 5^n$ for $n\geq 1$ ?
I don't understand the text in my textbook. I Would like somebody to explain it to me.

\begin{align}
a_{n+1} & = 6 \cdot 5^n + 2 a_n\\
& = 6 \cdot 5^n + 2 (6 \cdot 5^{n-1} + 2a_{n-1})\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2a_{n-1}\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2 (6 \cdot 5^{n-2} + 2a_{n-2})\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2 \cdot 6 \cdot 5^{n-2} + 2^3a_{n-2}\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2 \cdot 6 \cdot 5^{n-2} + 2^3(6 \cdot 5^{n-3} + 2 a_{n-3})\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2 \cdot 6 \cdot 5^{n-2} + 2^3 \cdot 6 \cdot 5^{n-3} + 2^4 \cdot a_{n-3}
\end{align}
Hence, in general (you need to prove the claim below using induction)
$$a_{n+1} = 2^{j+1}a_{n-j} + 6 \left( \sum_{k=0}^{j} 2^k 5^{n-k} \right)$$
Setting $j=n$, we get that
\begin{align}
a_{n+1} & = 2^{n+1}a_0 + 6 \left( \sum_{k=0}^{n} 2^k 5^{n-k} \right) = 2^{n+1}a_0 + 6 \cdot 5^n \left( \sum_{k=0}^{n} \left(\dfrac25 \right)^k \right)\\
& = 2^{n+1}a_0 + 6 \cdot 5^n \cdot \dfrac{1-(2/5)^{n+1}}{1-2/5}\\
& = 2^{n+1}a_0 + 6 \cdot \dfrac{5^{n+1} -2^{n+1}}3\\
& = 2^{n+1}a_0 + 2 \cdot (5^{n+1} -2^{n+1})
\end{align}
EDIT
You can verify if it is true by plugging the value for $a_n$ and $a_{n+1}$ and computing $a_{n+1} - 2a_n$.
$$a_n = 2^{n}a_0 + 2 \cdot (5^{n} -2^{n})$$
Hence, $$2a_n = 2^{n+1}a_0 + 4 \cdot (5^{n} -2^{n})$$
Hence,
\begin{align}
a_{n+1} - 2a_n & = 2^{n+1}a_0 + 2 \cdot (5^{n+1} -2^{n+1}) - \left( 2^{n+1}a_0 + 4 \cdot (5^{n} -2^{n})\right)\\
& = 2 \cdot (5^{n+1} -2^{n+1}) - 4 \cdot (5^n - 2^n)\\
& = 2 \cdot 5^{n+1} -2^{n+2} - 4 \cdot 5^n + 4 \cdot 2^n\\
& = 10 \cdot 5^{n} - 4 \cdot 5^n\\
& = 6 \cdot 5^n
\end{align}
