How many terms of the series 1 + 3 + 5 + 7 + amount to a sum equal...

The given series is $1 + 3 + 5 + 7 + \ldots$

This is an arithmetic progression with first term $a = 1$ and common difference $d = 2$.

The sum of the first $n$ terms of this series is given by:

$S_n = \dfrac{n}{2}[2a + (n-1)d] = \dfrac{n}{2}[2(1) + (n-1)2] = n^2$

We are given that $S_n = 12345678987654321$.

Therefore, $n^2 = 12345678987654321$.

Using the pattern of squares of numbers consisting entirely of ones:

$1^2 = 1$

$11^2 = 121$

$111^2 = 12321$

$\vdots$

\underbrace{11\ldots1}_{k \text{ times}^2 = 12\ldots k \ldots 21 (for $k \le 9$)

Since the maximum digit in the given sum is $9$, the square root consists of nine $1$s.

Thus, $n = 111111111$.

Answer: $111111111$