To find the sum of the given series, we can observe that each term is obtained by appending an additional 9 at the end and incrementing the preceding number by 1.
We can rewrite each term to express this pattern more clearly:
11 = 10 + 1
192 = 190 + 2
1993 = 1990 + 3
19994 = 19990 + 4
199995 = 199990 + 5
...
Notice that we can express each term of the series as follows:
11 = 10 + 1
192 = 100 + 90 + 2
1993 = 1000 + 900 + 90 + 3
19994 = 10000 + 9000 + 900 + 90 + 4
199995 = 100000 + 90000 + 9000 + 900 + 90 + 5
By breaking the terms into their respective place values (e.g., thousands, hundreds, tens, and ones), it becomes evident that we can simplify each term as follows:
11 = 10*1 + 1
192 = 100*1 + 90*1 + 2
1993 = 1000*1 + 900*1 + 90*1 + 3
19994 = 10000*1 + 9000*1 + 900*1 + 90*1 + 4
199995 = 100000*1 + 90000*1 + 9000*1 + 900*1 + 90*1 + 5
Now we notice a pattern: the term preceding the final number (in this case, 1999999999) has a coefficient of 1, the term before that has a coefficient of 10, the term before that has a coefficient of 100, and so on.
Using this pattern, we can express the sum of the series as:
Sum = (10*1 + 1) + (100*1 + 90*1 + 2) + (1000*1 + 900*1 + 90*1 + 3) + (10000*1 + 9000*1 + 900*1 + 90*1 + 4) + ...
Now, let's add the numbers in each place value separately:
Sum = (10 + 100 + 1000 + 10000 + ...) + (1 + 90 + 900 + 9000 + ...) + (1 + 2 + 3 + 4 + ...)
We can see that each set of numbers in parentheses forms a separate arithmetic series.
The first set, (10 + 100 + 1000 + 10000 + ...), is a geometric series with a common ratio of 10. We can use the formula for the sum of a geometric series to find its value:
Sum of geometric series = a * (r^n - 1) / (r - 1)
In this case, a = 10, r = 10, and n is the number of terms. Since we are summing terms up to 1999999999, n = 10.
Sum of the first set = 10 * (10^10 - 1) / (10 - 1)
Next, we calculate the sum of the second set, (1 + 90 + 900 + 9000 + ...), which is also a geometric series with a common ratio of 10. However, this series starts with 1 instead of 10.
Sum of the second set = 1 * (10^10 - 1) / (10 - 1)
Lastly, we find the sum of the third set, (1 + 2 + 3 + 4 + ...), which is a simple arithmetic series. We can use the formula for the sum of an arithmetic series to calculate it:
Sum of arithmetic series = (n/2) * (first term + last term)
In this case, n = 10 and the first term is 1, while the last term is 10.
Sum of the third set = (10/2) * (1 + 10)
Now, we add up the individual sums for each set to obtain the final answer:
Sum = [10 * (10^10 - 1) / (10 - 1)] + [1 * (10^10 - 1) / (10 - 1)] + [(10/2) * (1 + 10)]
Simplifying this expression will give us the actual sum of the series.