-formula.jpeg "1*2+2*3+3*4+...+n(n+1) Formula")
The Formula for 12 + 23 + 3*4 + ... + n(n+1)
The formula for the sum of the products of consecutive integers, namely 12 + 23 + 3*4 + ... + n(n+1), is a well-known result in mathematics. In this article, we will derive and explain this formula.
The Formula
The formula for the sum of the products of consecutive integers is given by:
$\sum_{k=1}^n k(k+1) = \frac{n(n+1)(n+2)}{3}$
Derivation
To derive this formula, we can start by writing out the sum explicitly:
$\sum_{k=1}^n k(k+1) = 12 + 23 + 3*4 + ... + n(n+1)$
We can rewrite this sum as:
$\sum_{k=1}^n k(k+1) = \sum_{k=1}^n (k^2 + k)$
Using the formula for the sum of squares and the formula for the sum of consecutive integers, we get:
$\sum_{k=1}^n k(k+1) = \sum_{k=1}^n k^2 + \sum_{k=1}^n k = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$
Simplifying this expression, we get:
$\sum_{k=1}^n k(k+1) = \frac{n(n+1)(n+2)}{3}$
Explanation
The formula for the sum of the products of consecutive integers is a quadratic function of n. This means that as n increases, the sum grows quadratically. This formula has many applications in mathematics and computer science, such as in the analysis of algorithms and the calculation of combinatorial quantities.
Conclusion
In this article, we have derived and explained the formula for the sum of the products of consecutive integers, namely 12 + 23 + 3*4 + ... + n(n+1). This formula is a useful result in mathematics and has many applications in various fields.
