Showing that numbers of the form 10101010...1 are composite

You have numbers of the form $\dfrac{100^n-1}{99}=\dfrac{(10^{n}-1)(10^{n}+1)}{9\times 11}$:

• If $n$ is odd you have $\dfrac{100^{2k-1}-1}{99}=\dfrac{10^{2k-1}-1}{9} \times \dfrac{10^{2k-1}+1}{11}$

• If $n$ is even you have $\dfrac{100^{2k}-1}{99}=\dfrac{10^{2k}-1}{99} \times \dfrac{10^{2k}+1}{1}$

and these factors are all integers, though some may be equal to $1$ when $k=1$

Hint: view the number as a geometric series. The first term is $1$ and the ratio is $100$. Use the formula for the sum of a geometric series, then see the numerator as a difference of two squares.