Divisibility by 101; a problem with induction

Question

I was trying to show that $10^{2n}+(-1)^{n+1}$ is divisible by $101$. Would anyone help me with the induction step please?

Answer 1

Hint. You should try writing out the sequence. For instance, for $n = 1$, we have $101$; for $n = 2$, we have $9999$; and so on. Then try dividing the second term by the first term, the third term by the second term, the fourth term by the third term. You should see a pattern that will help you construct the induction step. (It will be convenient to break it down by even and odd $n$, as hinted at by the $(-1)^{n+1}$.)

Answer 2

Taking module 101, $10^{2n}= 100^n\equiv (-1)^n(\mathrm{mod}101)$ and the result follows.

Answer 3

$$10^{2(n+1)}+(-1)^{(n+1)+1}=100\cdot10^{2n}-(-1)^{n+1}=101\cdot10^{2n}-\underbrace{(10^{2n}+(-1)^{n+1})}_{\text{multiple of }101}$$

Answer 4

Use the standard fact that $a-b \mid a^n-b^n$ as follows : $$101=100-(-1)\mid 100^n-(-1)^n=10^{2n}+(-1)^{n+1}$$

Answer 5

Notice the following steps of mathematical induction,

1. Setting $n=1$ in the given number $10^{2n}+(-1)^{n+1}$ gives $$10^{2\cdot 1}+(-1)^{1+1}=101$$ above number is divisible by $101$ for $n=1$

2. assume that $10^{2n}+(-1)^{n+1}$ is divisible by $101$ for $n=k$ then $$10^{2k}+(-1)^{k+1}=101m$$ or $$10^{2k}=101m-(-1)^{k+1}\tag 1$$ where, $m$ is some integer

3. setting $n=k+1$, $$10^{2(k+1)}+(-1)^{k+1+1}$$ $$=100\cdot 10^{2k}-(-1)^{k+1}$$ setting the value of $10^{2k}$ from (1), $$=100(101m-(-1)^{k+1})-(-1)^{k+1}$$ $$=101(100m)-101(-1)^{k+1}$$ $$=101(100m-(-1)^{k+1})$$

since, $(100m-(-1)^{k+1})$ is an integer hence the number $101(100m-(-1)^{k+1})$ is divisible by $101$

thus $10^{2n}+(-1)^{n+1}$ is divisible by $101$ for $n=k+1$

hence, $10^{2n}+(-1)^{n+1}$ is divisible by $101$ for all integers $n\ge 1$