Sum of 1 to n using Mathematical Ascending Method

The question asks for the value of the sum \(1 + 2 + 3 + \ldots + n\) using the mathematical ascending method. This method essentially refers to finding the sum of an arithmetic progression where the first term is 1, the last term is \(n\), and the common difference is 1. There are \(n\) terms in this sequence.

Understanding the Sum of an Arithmetic Series

The sum of an arithmetic series can be found using a well-known formula. If the series has \(N\) terms, the first term is \(a_1\), and the last term is \(a_N\), the sum \(S_N\) is given by:

\(S_N = \frac{N}{2} (a_1 + a_N)\)

Applying the Formula to 1 + 2 + ... + n

For the series \(1 + 2 + 3 + \ldots + n\):

• The number of terms \(N\) is \(n\).
• The first term \(a_1\) is 1.
• The last term \(a_N\) is \(n\).

Substituting these values into the sum formula:

\(S_n = \frac{n}{2} (1 + n)\)

This can also be written as:

\(S_n = \frac{n(n + 1)}{2}\)

This formula gives the sum of the first \(n\) natural numbers, which is the value of \(1+2+3+\ldots+n\) by the mathematical ascending method.

Comparing with Options

Let's compare the derived formula \(\frac{n(n + 1)}{2}\) with the given options:

• Option 1: \(\frac{(n + 1)(n + 2)}{2}\)
• Option 2: \(\frac{(n + 1)(n + 2)}{3}\)
• Option 3: \(\frac{n(n + 1)}{3}\)
• Option 4: \(\frac{n(n + 1)}{2}\)

The formula we derived, \(\frac{n(n + 1)}{2}\), matches Option 4.

Therefore, the value of \(1+2+3+\ldots+n\) by the mathematical ascending method is \(\frac{n(n + 1)}{2}\).