Prove that 2.101001000100001... is an irrational number.

Let $x = 2 + \sum\limits_{k=1}^\infty10^{-k(k+1)/2}$ be the number at hand. If $x$ is rational, say $x = \frac{p}{q}$ for some positive integers $p,q$, we can pick a $n > 1$ such that $10^n > q + 1$. It is clear $$qx \times 10^{n(n-1)/2} = p \times 10^{n(n-1)/2}$$ is also an integer. However, the fractional part of this number is equal to

$$ \left\{ q \times 10^{n(n-1)/2} \left(2 + \sum_{k=1}^\infty 10^{-k(k+1)/2}\right)\right\} = \left\{ q \times \sum_{k=1}^\infty 10^{-k(k+2n-1)/2} \right\} $$ which belongs to $(q \times 10^{-n}, (q+1)\times 10^{-n} ) \subset (0,1)$. Since $(0,1)$ doesn't contain any integer, this leads to a contradiction and hence $x$ is irrational.

Hint:

2 is rational
$0.1010010001\ldots$ is easy to show irrational (proof in first chapter of Rudin IIRC)

rational + irrational = irrational

You have $$x=\sum_{n=0}^{\infty}10^{-\binom{n}{2}}$$

If you have a repeating decimal then you have finitely many digits appearing first, followed by strings of any length that repeat (any number of times) later in the representation. This decimal representation is not repeating, since for any $k$, we can identify a string of digits $1\overbrace{00\cdots00}^k1$ that is present in the decimal representation and never appears again. If all such strings were present in the nonrepeating part, we have a contradiction, since the nonrepeating part is only finitely many digits long.