Αποτελέσματα Αναζήτησης
Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers. It is done in two steps. The first step, known as the base case, is to prove the given statement for the first natural number.
Definition: Mathematical Induction. To show that a propositional function is true for all integers , follow these steps: Basis Step: Verify that is true. Inductive Step: Show that if is true for some integer , then is also true. The basis step is also called the anchor step or the initial step.
I need to prove that $$ 2^n > n^3\quad \forall n\in \mathbb N, \;n>9.$$ Now that is actually very easy if we prove it for real numbers using calculus. But I need a proof that uses mathematical induction.
13 Ιουν 2020 · By the distributive law, $(x + 1)(x - 1) = x^2 - 1$. Set $x = -1$ and observe that the left hand side is zero, so the right hand side is zero, so $x^2 = 1$. It's arguable whether the assumptions I use there are any more or less obvious than the conclusion.
17 Απρ 2022 · Use mathematical induction to prove each of the following: (a) For each natural number \ (n\), 3 divides \ ( (4^n - 1)\). (b) For each natural number \ (n\), 6 divides \ ( (n^3 - n)\). In Exercise (7), we proved that for each natural number \ (n\), \ (4^n \equiv 1\) (mod 3).
Show how to prove if a sum of infinite terms diverges or converges with different tests: limit test, ratio test, root test, integral test, p-series test or geometric series test. Investigate the convergence or divergence of an infinite sum step by step:
Solution. Verified by Toppr. Given series is $$9^ {1/3}\times 9^ {1/9}\times 9^ {1/27}\times .....\infty$$ $$=9^ {\left (\tfrac {1} {3}+\tfrac {1} {9}+\tfrac {1} {27}+....\infty\right)}=9^ {\tfrac {a} { (1-r)}}$$, where $$a=\dfrac {1} {3}$$ and $$r=\dfrac {1} {3}$$
