Answer
See below.
Work Step by Step
1. Let $P(n)$ be the statement to be proved.
2. Test for $n=0,1,2$, we have $e_0=5(3^0)+7(2^0)=12, e_1=5(3^1)+7(2^1)=29, e_2=5e_1-6e_0=5(29)-6(12)=73$ while with the formula $e_2=5(3^2)+7(2^2)=73$, thus $ P(0),P(1),P(2)$ are all true.
3. Suppose it is true for $n\le p, (p\gt2)$, that is $e_p=5(3^p)+7(2^p), e_{p-1}=5(3^{p-1})+7(2^{p-1})...$
4. For $n=p+1$, we have $e_{p+1}=5e_p-6_{p-1}=5(5(3^p)+7(2^p))-6(5(3^{p-1})+7(2^{p-1}))=25(3^p)+35(2^p)-10(3^p)-21(2^p)=15(3^p)+14(2^p)=5(3^{p+1})+7(2^{p+1})$ confirming the formula.
5. Thus $P(p+1)$ will also be true and we have proved the statement using mathematical induction.