The given series is \(1+3+5+7+...\). This is an arithmetic series where:
We need to find the number of terms (\(n\)) such that the sum of these terms (\(S_n\)) equals \(12345678987654321\).
The formula for the sum of the first \(n\) terms of an arithmetic series is:
\(S_n = \frac{n}{2} [2a + (n-1)d]\)
Substitute the values \(a=1\) and \(d=2\) into the formula:
\(S_n = \frac{n}{2} [2(1) + (n-1)2]\)
\(S_n = \frac{n}{2} [2 + 2n - 2]\)
\(S_n = \frac{n}{2} [2n]\)
\(S_n = n^2\)
This shows that the sum of the first \(n\) odd numbers is equal to the square of \(n\). In this specific series, the sum of the first \(n\) terms is \(n^2\).
We are given that the sum \(S_n = 12345678987654321\). Therefore, we have the equation:
\(n^2 = 12345678987654321\)
To find \(n\), we need to calculate the square root of \(12345678987654321\).
Let's examine the pattern of squares of numbers consisting of ones:
| Number | Square | Pattern |
| \(1\) | \(1^2 = 1\) | \(1\) |
| \(11\) | \(11^2 = 121\) | \(121\) |
| \(111\) | \(111^2 = 12321\) | \(12321\) |
| \(1111\) | \(1111^2 = 1234321\) | \(1234321\) |
| \(111111111\) | \(111111111^2 = 12345678987654321\) | \(12345678987654321\) |
By observing the pattern, we can see that the square root of \(12345678987654321\) is \(111111111\).
So, \(n = \sqrt{12345678987654321} = 111111111\)
The number of terms of the series \(1+3+5+7+...\) that amount to a sum equal to \(12345678987654321\) is \(111111111\).
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