anemone said:Prove $\dfrac{1}{3}\cdot\dfrac{4}{6}\cdot\dfrac{7}{9}\cdots\dfrac{1000}{1002}\lt \dfrac{1}{79}$
anemone said:Prove $\dfrac{1}{3}\cdot\dfrac{4}{6}\cdot\dfrac{7}{9}\cdots\dfrac{1000}{1002}\lt \dfrac{1}{79}$
The proof involves using mathematical induction, where we first prove that the statement is true for the base case (n=1) and then show that if it is true for any arbitrary integer k, then it must also be true for the next integer (k+1). By repeating this process, we can prove that the statement holds for all positive integers n, including n=1000.
An algebraic series is a sequence of numbers that follow a specific pattern or rule. Each term in the series is obtained by performing a mathematical operation on the previous term. For example, the sequence 1, 3, 5, 7, 9 is an algebraic series where each term is obtained by adding 2 to the previous term.
Proving this statement can have applications in various areas of mathematics, such as number theory and analysis. It demonstrates the power of mathematical induction as a proof technique and also helps us understand the behavior of algebraic series.
Mathematical induction is a proof technique used to prove statements about all positive integers. It involves proving a base case (usually n=1) and showing that if the statement is true for any arbitrary integer k, then it must also be true for the next integer (k+1). By repeating this process, we can prove that the statement holds for all positive integers n.
Yes, the proof for the 1000th progression of an algebraic series being less than 1/79 can be extended to other algebraic series. However, the specific values and operations may differ for each series, so the proof would need to be adapted accordingly.