'Mathematical Induction'

Question

Use mathematical induction to prove that $4^n -3^n + 1 = 7a_{n-1} – 12a_{n-2} + 6$ with $n \ge 3$ with the initial condition $a_1 = 2$ and $a_2 = 8$ . Given that $a_n = 4^n -3^n + 1$.

I am confused at the inductive step or $(k+1)$ part.

Answer 1

Since the OP asks for an inductive proof, I'll write in this format.

Base case

• $n = 3$
  • $\text{LHS} = 4^3 - 3^3 + 1 = 64 - 27 + 1 = 38$
  • $\text{RHS} = 7a_2 - 12a_1 + 6 = 7(8) - 12(2) + 6 = 56 - 24 + 6 = 38$
• $n = 4$
  • $\text{LHS} = 4^4 - 3^4 + 1 = 256 - 81 + 1 = 176$
  • $\text{RHS} = 7 a_3 - 12 a_2 + 6 = 7(38) - 12(8) + 6 = 266 - 96 + 6 = 176$

Induction hypothesis (redundant)

• $n = k-1, a_{k-1} := 4^{k-1} - 3^{k-1} + 1 = 7 a_{k-2} - 12 a_{k-3} + 6$
• $n = k, a_k := 4^k - 3^k + 1 = 7 a_{k-1} - 12 a_{k-2} + 6$

Inductive step

\begin{align*} n &= k+1 \\ a_{k+1} &:= 4^{k+1} - 3^{k+1} + 1 \\ &= 4 \cdot 4^k - 3 \cdot 3^k + 1 \\ &= 7 (4^k - 3^k + 1) - 3 \cdot 4^k + 4 \cdot 3^k - 6 \\ &= 7 a_k - 12 (4^{k-1} - 3^{k-1} + 1) + 6 \\ &= 7 a_k - 12 a_{k-1} + 6 \end{align*}

Hence, by induction, the proposition is true for all $n \in \Bbb N$.

Remarks

The induction hypothesis is not needed. As vrugtehagel says, what we're doing is just verifying $4^n-3^n+1=7a_{n-1}-12a_{n-2}+6 \,\forall\,n\ge3$.