I am struggling with a math problem I have been assigned. The problem is as follows:
Let $X_1 = -3$ and $X_2 = 0$. Given that for every natural number $n \geq 2, X_{n+1} = 7X_n - 10X_{n-1}$, prove by induction that for every $n$ belonging to $\mathbb{N}$, $X_n = 2 \cdot 5^{n-1} - 5 \cdot 2^{n-1}$
Right now all I have proven so far is the two base cases, I am not sure how to take this proof any further. Any assistance you can offer is greatly appreciated, thank you.
Hint: Use strong induction (i.e. you know it is true for all $k<n$ now prove it for $n$).
$$X_{n} = 7X_{n-1}-10X_{n-2} = 7(2\cdot 5^{n-2}-5\cdot 2^{n-2})-10(2\cdot 5^{n-3} -5\cdot 2^{n-3})$$
Then, simplify the right hand side and see if you get what you need...
Start with your recurrence relation and substitute your inductive assumption:
$$X_{n+1} = 7(2 \cdot 5^{n-1} - 5 \cdot 2^{n-1}) - 10(2 \cdot 5^{n-2} - 5 \cdot 2^{n-2}) \\ = 14 \cdot 5^{n-1} - 35 \cdot 2^{n-1} - 20 \cdot 5^{n-2} + 50 \cdot 2^{n-2} \\ = 10 \cdot 5^{n-1} - 10 \cdot 2^{n-1} \\ = 2 \cdot 5^n - 5 \cdot 2^n.$$
Assume true for $n = k$, Which means
${X_k} = 2 \cdot {5^{k - 1}} - 5 \cdot {2^{k - 1}}$
${X_{k + 1}} = 7{X_k} - 10{X_{k - 1}}$
Required to prove for $n = k + 1$ That is required to prove
${X_{k + 1}} = 2 \cdot {5^k} - 5 \cdot {2^k} = 7(2 \cdot {5^{k - 1}} - 5 \cdot {2^{k - 1}}) - 10(2 \cdot {5^{k - 2}} - 5 \cdot {2^{k - 2)}})$
or
$2 \cdot {5^k} - 5 \cdot {2^k} - 7(2 \cdot {5^{k - 1}} - 5 \cdot {2^{k - 1}}) + 10(2 \cdot {5^{k - 2}} - 5 \cdot {2^{k - 2)}}) = 0$
Now
$2 \cdot {5^k} - 5 \cdot {2^k} - 7(2 \cdot {5^{k - 1}} - 5 \cdot {2^{k - 1}}) + 10(2 \cdot {5^{k - 2}} - 5 \cdot {2^{k - 2)}}) =$
$ = 2 \cdot {5^k} - 14 \cdot {5^{k - 1}} + 20 \cdot {5^{k - 2}} - 5 \cdot {2^k} + 35 \cdot {2^{k - 1}} - 50 \cdot {2^{k - 2}}$
$ = 2 \cdot {5^{k - 2}}({5^2} - 7 \times 5 + 10) + 5 \cdot {2^{k - 2}}( - {2^2} + 7 \times 2 - 10)$
$ = 2 \cdot {5^{k - 2}}(0) + 5 \cdot {2^{k - 2}}(0)$
$ = 0$
