Proof that all terms of a sequence end in 7

Question

The terms of a sequence are given by the equation $T_{n} = 5 \times 3^{n-1} + 2$.

$5 \times 3^{n-1}$ will always end in 5 since it is divisible by 5 but not 2

So $T_{n}$ will always end in 7.

Is there another, more rigorous way, to prove that all terms of the sequence will end in 7?

Answer 1

Proof by induction!

Theorem For all $n \in \mathbf{N}$, $T_n = 10a + 7$, with $a \in \mathbf{N}$.

Base case $n=1$ gives $T_n = 7$.

Induction step Let $n = k+1$. Then,

\begin{align} T_{k+1} & = 5 \times 3^k + 2 = 3 \times 5 \times 3^{k-1} + 2 \\ & = 3 (T_k-2)+2 = 3(10a+5)+2 = 10a^\prime +7, \end{align} with $a^\prime \in \mathbf{N}$.

Answer 2

$$3\equiv1\pmod2\implies5\cdot3^m\equiv5\pmod{5\cdot2}$$ for any positive integer $m$

$$\implies5\cdot3^m+2\equiv5+2$$