Calculating Infinite Series with Repeating Ones: A Deep Dive

Understanding the sum of infinite series, particularly those consisting of repeating digits, can be fascinating and insightful. In this article, we explore a specific series featuring repeating ones and provide a detailed method for calculating its sum. We will also cover the formula derivation, its application, and related examples for clarity.

Introduction to the Series

Consider the series:

1 11 111 1111 11111 ...

Each term in the series is a number consisting of a sequence of ones. For instance, the n-th term can be represented as a series of n ones. This can be expressed mathematically as:

[ T_n frac{10^n - 1}{9} ]

This formula is derived from the geometric series representation of the repeating ones. In this section, we will explore the derivation of this formula and its application in calculating the sum of the series.

The Sum of the Series

To find the sum of the first n terms of the series, we start by expressing the sum of the first n terms:

[ S_n T_1 T_2 T_3 ldots T_n ]

Using the derived formula for each term T_n, we have:

[ S_n sum_{k1}^{n} frac{10^k - 1}{9} ]

We can separate the sums for easier calculation:

[ S_n frac{1}{9} sum_{k1}^{n} 10^k - frac{1}{9} sum_{k1}^{n} 1 ]

The first sum is a geometric series, and the second sum is simply the sum of n ones:

[ sum_{k1}^{n} 10^k 10 frac{10^n - 1}{10 - 1} frac{10^{n 1} - 10}{9} ] [ sum_{k1}^{n} 1 n ]

Substituting these back into the equation for S_n:

[ S_n frac{1}{9} left( frac{10^{n 1} - 10}{9} - n right) ]

Combining these, we get the final formula for the sum of the first n terms:

[ S_n frac{10^{n 1} - 10 - 9n}{81} ]

Application and Examples

Now let's apply this formula to solve for S in a given problem:

Example 1

In this example, we are given:

[ 12345 frac{10^{15} - 10 - 9n}{81} ]

We need to find n such that the sum is 12345:

[ 12345 frac{10^{15} - 10 - 9n}{81} ]

Multiplying both sides by 81:

[ 12345 times 81 10^{15} - 10 - 9n ]

Simplifying:

[ 100000 10^{15} - 10 - 9n ]

Since (10^{15}) is significantly larger, we can approximate the terms:

[ 100000 approx 10^{15} - 10 ]

Thus, we can solve for n:

[ 9n approx 10^{15} - 100000 - 10 ]

[ 9n approx 10^{15} - 100010 ]

[ n approx frac{10^{15} - 100010}{9} ]

This shows how to solve for n given a specific sum.

Example 2

Now, consider the more intricate problem of finding the sum:

[ S 1 11 111 1111 11111 ldots ]

Let's find the sum:

[ S 1 10 100 1000 10000 ldots ]

Using the formula for the sum of the first n terms:

[ S frac{10^{n 1} - 10 - 9n}{81} ]

For practical purposes, let's consider the sum of the first 7 terms:

[ S 1234567 ]

This calculation confirms that the sum for the first 7 terms is 1234567.

Conclusion

In summary, the sum of the series consisting of repeating ones can be calculated using the formula: