prove that $(k+1)^a - 1$ is divisible by k
Base case: n = 0;
$(k+1)^0$ -1 = k which is divisible by k
IH: $(b+1)^a - 1$ is divisible by b
for a + 1
$(b+1)^{a+1}$ -1 = $(b+1)(b+1)^{a} -1 $
Asked
Active
Viewed 48 times
1 Answers
1
From what you have above, I assume that $k$, $a$, and $b$ are all natural numbers.
Then you can say (IH): $$(b+1)^{a}-1=bm$$ where $m$ is also a real number. Now for $a+1$: $$(b+1)^{a+1}-1$$ $$\Rightarrow (b+1)(b+1)^{a}-1$$ $$\Rightarrow b(b+1)^{a}+(b+1)^{a}-1$$ Now you can make the substitution for the (IH) $$\Rightarrow b(b+1)^{a}+bm$$ $$\Rightarrow b((b+1)^{a}+m)$$ Hence, since $(b+1)^{a}+m)$ is a natural number (call it $n$), you can write this final proof as $bn$, which is obviously divisible by $b$.
