(a) $5$ $|$ $3^{3n+1}+2^{n+1}$
(b) $21$ $|$ $4^{n+1} + 5^{2n-1}$
(c) $24$ $|$ $2 \cdot7^n + 3 \cdot5^n - 5$
These are trivial by using induction. But I have tried to prove it by binomial theorem and remainder theorem, but couldn't succeed. Please help to prove it.
Hints with modular arithmetic:
$$(a)\;\;3^3=2\pmod 5\implies \left(3^{3}\right)^n\cdot 3=2^n\cdot 3\implies 3^{3n+1}+2^{n+1}=2^n(3+2)\pmod 5$$
$$(b)\;\;5^2=4\pmod {21}\implies 4^{n+1}+5^{2n-1}=4^n(4+5^{-1})\pmod{21}\;,\;$$
$$\text{but}\;5^{-1}=-4\pmod{21}\ldots$$
Now you try something similar as the above for the third one
(c)
Method $1:$ Observe that $7^2=49\equiv 1\pmod{24}$ and $5^2=25\equiv 1\pmod{24}$
So, when $n$ is even, $2(7^n)+3(5^n)\equiv2+3\pmod{24}\equiv5$
If $n$ is odd, $2(7^n)+3(5^n)\equiv2\cdot7+3\cdot5\pmod{24}\equiv29\equiv5$
Method $2:$
$\displaystyle2(7^n)+3(5^n)=2(1+6)^n+3(1+4)^n$
$\displaystyle=2\{1+6n+6^2(\text{ some integer })\}+3\{1+4n+4^2(\text{ some integer })\}$ (using Binomial Expansion of positive integer index)
$\displaystyle\equiv 2+12n+3+12n\pmod{24}\equiv5$
(b) $$4^{n+1}+5^{2n-1}=16(4^{n-1})+5(5^{2n-2})=16(4^{n-1})+5(25^{n-1})$$ $$\equiv16(4^{n-1})+5(4^{n-1})\pmod {21}\text{ as }25\equiv4\pmod {21}$$
$\implies 4^{n+1}+5^{2n-1}\equiv4^{n-1}(16+5)\pmod{21}$
(a) $3^{3n+1}+2^{n+1}=3(27^n)+2(2^n)\equiv 3(2^n)+2(2^n)\pmod 5\equiv 5(2^n)$
