1. 1 + 2 + 3 + ··· + n = n(n + 1)(2n + 1) 6 Proof: For n = 1 the
In Exercises 1-15 use mathematical induction to establish the formula for n ? 1. 1. 1. 2. + 2. 2. + 3. 2. + ··· + n. 2. = n(n + 1)(2n + 1).
MATHEMATICAL INDUCTION SEQUENCES and SERIES
?n is always divisible by 30. Use the fact that n. 5. ?n = (n?1)n(n+1)(n. 2. +1) to prove that it is divisible by 2 and 3 as well as 5.
Problem: For each positive integer n the formula 1 · 3+2 · 4+3 · 5 +
1 · 3+2 · 4+3 · 5 + ··· + n(n + 2) = n(n + 1)(2n + 7). 6 is valid. Proof: (formal style; it is good to do a few proofs this way) We will use the Principle
Proof by Induction - University of Plymouth
12-Feb-2006 Example 3: for n a natural number prove that: 1) if n ? 2 then n3 ? n is always divisible by 3
X X
2! + t4. 4! 3 t6. 6! + 17 t8. 8!
Problem Solving for Math Competitions Harm Derksen
* Prove that. 12 + 22 + ··· + n2 = n(n + 1)(2n + 1). 6 for all positive integers n. Exercise 1.2. * Show that. 1 ?. 1. 2. +. 1. 3. ?
BINOMIAL THEOREM
8.1.3 Some important observations. 1. The total number of terms in the binomial expansion of (a + b)n is n + 1 i.e. one more than the exponent n. 2.
CS240 Solutions to Induction Problems Fall 2009 1. Let P(n) be the
If we had shown P(3) as our basis step then the inequality would only be proven for n ? 3. 2. For any positive integer n n. ? i=1. 1 i(i + 1). = 1. 1 · 2.
IMO 2008 Shortlisted Problems
(a ? 1)(b ? 1)(c ? 1) = abc. ?=(a ? 1)2 ? 4a(a ? 1) = (1 ? a)(1 + 3a). ... (i) f(2n ? t(m)) ? 3(n?1)/2 if 2n + m is divisible by 3;.
Untitled
principle of induction P(n) is true for all positive integers n > 1. herefore each term in the product (1+2+22)(1 +3 +3² +3³)(1 +.
Polynomials II - Divisibility and Irreducibles
So now for instance x2 ?2 is not divisible by x? ? 2 over Q since x? ? 2 doesn’t exist over Q Problem 9 Show that x2?2 is irreducible over Q (Hint: Suppose it were reducible What would the degrees of the factors have to be and what does that mean?) Problem 10 Show that x3 ?2 is irreducible over Q (Hint: This is similar
Math 115 Exam Solutions - Colorado State University
1+n2 = 0 and (ii) the sequence of terms 1+n2 are decreasing To see (i) notice that we can divide numerator and denominator by n2 to get lim n?? 1 n2 ·n 1 n2 (1+n 2) = lim n?? 1 n 1 n2 +1 = 0 To see (ii) let f(x) = x 1+x2 Then f0(x) = (1+x2)·1?x·2x (1+x2)2 = 1?x2 (1+x2)2 ? 0 for x ? 1 Therefore f is a decreasing
3 Mathematical Induction 31 First Principle of
3 MATHEMATICAL INDUCTION 89 Which shows 5(n+ 1) + 5 (n+ 1)2 By the principle of mathematical induction it follows that 5n+ 5 n2 for all integers n 6 Discussion In Example 3 4 1 the predicate P(n) is 5n+5 n2 and the universe of discourse
How to prove n is true for all integers n 1?
Mathematical induction can be used to prove that a statement about n is true for all integers n ? 1. We have to complete three steps. In the basis step, verify the statement for n = 1. In the inductive hypothesis, assume that the statement holds when n = k for some integer k ? 1.
Does f(n) = n 1+n2 converge?
Therefore, f is a decreasing function in the relevant range, so the terms f(n) =n 1+n2are decreasing. 1 We know that the series converges, but we need to determine whether it converges absolutely or not. In other words, we must determine if X? n=1 (?1) nn 1+n2 = X? n=1 n 1+n2 . converges or not.
Is p divisible by Q over C?
Definition 1Let p and q be polynomials in the complex numbers C. We say that p is divisble by q over C if there exists a polynomial r such that p(z) = q(z)r(z). Problem 1 By long division (which we learned how to do last week), answer the following: • Is x3?2x2+ x?2 divisible by x?2 over C?
