4.6 Arithmetic Sequences 20, 25, 30, 35
... Arithmetic Sequences definition: a sequence created by adding the same number repeatedly. ...
... Arithmetic Sequences definition: a sequence created by adding the same number repeatedly. ...
1 Prime Numbers The natural numbers, or the counting numbers, 1
... Let us use this method of proof to show that there are infinitely many primes. We assume that there are a finite number of primes. Let us denote these primes by p1 , p 2 , p3 ,..., p n . We form the number N = p1 × p 2 × p3 × ... × p n +1. Now N is a natural number greater than 1; so N can be class ...
... Let us use this method of proof to show that there are infinitely many primes. We assume that there are a finite number of primes. Let us denote these primes by p1 , p 2 , p3 ,..., p n . We form the number N = p1 × p 2 × p3 × ... × p n +1. Now N is a natural number greater than 1; so N can be class ...
CA ADV Algebra Standard 06
... First, expand the square and multiply. Then, combine like terms and simplify. i squared is equal to –1. First, expand the square and multiply. Then, combine like terms and simplify. Correct! ...
... First, expand the square and multiply. Then, combine like terms and simplify. i squared is equal to –1. First, expand the square and multiply. Then, combine like terms and simplify. Correct! ...
Full text
... the Golden Ratio. It is well-known that Φ ≈ 1.6180339 has the unique property that Φ and Φ−1 have the same decimal part since Φ = 1 + Φ−1 . Also, if {Hn } satisfies the Fibonacci recursion with H1 = a, H2 = b, then Hn = aFn−1 + bFn−2 and the ratio Hn+1 /Hn also approaches Φ as a limit [1, 2, 3, 4]. ...
... the Golden Ratio. It is well-known that Φ ≈ 1.6180339 has the unique property that Φ and Φ−1 have the same decimal part since Φ = 1 + Φ−1 . Also, if {Hn } satisfies the Fibonacci recursion with H1 = a, H2 = b, then Hn = aFn−1 + bFn−2 and the ratio Hn+1 /Hn also approaches Φ as a limit [1, 2, 3, 4]. ...
Methods of Proof
... n is even. Then we can express n as 2k, where k is an integer. Therefore 3n+2 is then 6k+2, i.e. 2(3k+1), and this is an even number. This contradicts our assumptions, consequently n must be odd. Therefore when 3n+2 is odd, n is odd. QED ...
... n is even. Then we can express n as 2k, where k is an integer. Therefore 3n+2 is then 6k+2, i.e. 2(3k+1), and this is an even number. This contradicts our assumptions, consequently n must be odd. Therefore when 3n+2 is odd, n is odd. QED ...
