Understanding the Arithmetic Series
The given series is \(1+3+5+7+...\). This is an arithmetic series where:
- The first term (\(a\)) is \(1\).
- The common difference (\(d\)) is \(3 - 1 = 2\).
We need to find the number of terms (\(n\)) such that the sum of these terms (\(S_n\)) equals \(12345678987654321\).
Calculating the Sum of the Series
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 \(n^2\).
Finding the Number of Terms
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, \(n = \sqrt{12345678987654321} = 111111111\)
Conclusion
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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Important Questions from Sequences and Series
The sum of the first three terms of an arithmetic progression (A.P.) is $24$, and the sum of its next three terms (i.e., the $4^{th}$, $5^{th}$, and $6^{th}$ terms) is $51$. What is the sum of the first $10$ terms of this A.P.?
What is the limit point of the sequence < f > = 1?
The sequence given by interval [0,1] is ______.
Find the limit point of the sequence <1, 2, 1/2, 3, 1/3..... >
- \(\mathop {\lim }\limits_{n \to \infty } {\left( {1 - \frac{1}{{2n}}} \right)^{n + 1}}\) is equal to
