Question

Prove that \(1 \cdot 2+2 \cdot 3+\cdots+n \cdot(n+1)=n(n+1)(n+2) / 3 .\) This problem should not be solved using a proof by induction.

Short answer

The formula is verified: \(1 \cdot 2 + 2 \cdot 3 + \cdots + n \cdot (n+1) = \frac{n(n+1)(n+2)}{3}\).

Step-by-step solution

Identify the Series Structure

The given series is \(1 \cdot 2 + 2 \cdot 3 + \cdots + n \cdot (n+1)\). Each term in the series can be expressed in the form \(k(k+1)\), where \(k\) ranges from 1 to \(n\). We need to find a closed-form formula for this series that matches the right-hand side of the given equation.

Expand the General Term

Let's expand the general term \(k(k+1)\) as \(k^2 + k\). Thus, the series can be rewritten as \(\sum_{k=1}^{n} (k^2 + k)\). This becomes two separate sums: \(\sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k\).

Use Sum Formulas

Recall the formulas for the sum of the first \(n\) squares and the sum of the first \(n\) integers:- \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\)- \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\)Using these, the series becomes:\[\sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\]

Simplify the Expression

Combine the two fractions:\(\frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6}\) since \(\frac{n(n+1)}{2} = \frac{3n(n+1)}{6}\).Combine these over a common denominator:\[\frac{n(n+1)(2n+1) + 3n(n+1)}{6}\]

Factor and Simplify Further

Simplify inside the numerator:\(n(n+1)(2n+1) + 3n(n+1) = n(n+1)(2n+1 + 3) = n(n+1)(2n+1 + 3)\).This simplifies to:\(n(n+1)(2n+4)\).Factor out a 2 from \(2n+4\):\(n(n+1)(2)(n+2)\).

Complete the Proof

Continue to factor such that it shows:\(\frac{2n(n+1)(n+2)}{6} = \frac{n(n+1)(n+2)}{3}\), as we retrieve the formula in the problem statement.Thus, the series \(1 \cdot 2 + 2 \cdot 3 + \cdots + n \cdot (n+1) = \frac{n(n+1)(n+2)}{3}\) is verified.

Key concepts

Algebraic manipulation

Algebraic manipulation involves using algebraic techniques to rearrange and simplify expressions. In the context of the given problem, we're presented with a series that appears complex at first glance: the sum of terms of the form \( k(k+1) \).

The first step in working with such expressions is to expand each term, which helps to see the structure more clearly. By expanding \( k(k+1) \) to \( k^2 + k \), we break it down into simpler components. This process highlights the utility of algebraic manipulation — turning a complex expression into simpler parts that are easier to handle.

This breakdown sets us up to use known formulas for summing series, specifically the sum of squares and the sum of natural numbers. Through methodical reduction and combination of terms using algebra, it's possible to transform complex series into manageable forms ready for further simplification or computation.

Closed-form expression

A closed-form expression is a way of representing a sequence or series by a single formula, as opposed to an infinite sum. For any given sequence, a closed-form expression provides a straightforward evaluation method, without needing to compute every individual term.

In this problem, we're tasked with converting the series \( 1 \cdot 2 + 2 \cdot 3 + \cdots + n \cdot (n+1) \) into a closed-form expression. By employing known summation formulas for \( \sum_{k=1}^{n} k^2 \) and \( \sum_{k=1}^{n} k \), and then combining these through algebraic manipulation, we find that the closed form, \( \frac{n(n+1)(n+2)}{3} \), succinctly represents the entire sequence.

The beauty of a closed-form expression lies in its efficiency. It allows you to calculate the sum for any given \( n \) directly, without summing up each term individually. This significantly reduces computation time, especially for large values of \( n \), and demonstrates why finding closed-form expressions is a powerful tool in mathematics.

Series transformation

Series transformation is the process of converting one series form into another, often to simplify or find a solution. In this exercise, we started with a straightforward, albeit cumbersome, series involving products: \( 1 \cdot 2, 2 \cdot 3, \ldots, n \cdot (n+1) \).

The key transformation here was to express each product as \( k(k+1) \) and then separate into simpler sums: \( k^2 \) and \( k \). This transformation helps in applying known summation formulas, allowing the original series to be expressed as a sum of simpler series that we can handle more easily.

Furthermore, with a common denominator, both terms \( \frac{n(n+1)(2n+1)}{6} \) and \( \frac{3n(n+1)}{6} \) are combined and transformed into a single expression. The result, \( \frac{n(n+1)(n+2)}{3} \), reflects how an initial series can undergo systematic transformation to become more tractable.

Ultimately, the transformation techniques make the process of solving, understanding, and utilizing series more approachable, converting complex arithmetic series into elegant mathematical formulas.