1. a) Deduce from the relation Fk+1F-1-F2 = (-1)k, that for any positive integer k > 2 Fk b) Deduce from part a) that for any positive integer n, Fn+1 = Σ 1)k+1
1. a) Deduce from the relation Fk+1F-1-F2 = (-1)k, that for any positive integer k > 2 Fk b) Deduce from part a) that for any positive integer n, Fn+1 = Σ 1)k+1
DEFINITION: For a positive integer n, τ(n) is the number of
positive divisors of n and σ(n) is the sum of those divisors.
4. The goal of this problem is to prove the inequality in part (b), that o(1)+(2)+...+on) < nº for each positive integer n. The first part is a stepping-stone for that. (a) (10 points.) Fix positive integers n and k with 1 <ksn. (i) For which integers i with 1 <i<n is k a term in the...
-870 (1 point) If n is a positive integer, then integer, then [8 1739]" (1 2 (Hint: Diagonalize the matrix [i 19 ) mest. Note that your ansı (Hint: Diagonalize the matrix first. Note that your answer will be a formula that involves n. Be careful with parentheses.)
Q18 12 Points For any positive integer n, let bn denote the number of n-digit positive integers whose digits are all 1 or 2, and have no two consecutive digits of 1. For example, for n - 3, 121 is one such integer, but 211 is not, since it has two consecutive 1 's at the end. Find a recursive formula for the sequence {bn}. You have to fully prove your answer.
Find σ. (Enter an exact number as an integer, fraction, or decimal.) X ~ N(3, 5)
3. Find a closed formula for the exponential generating function A(x) Σ an,n wh n+1-(n + 1)(m-n + 1), a,-1. ere an satisty the recursion a
3. Find a closed formula for the exponential generating function A(x) Σ an,n wh n+1-(n + 1)(m-n + 1), a,-1. ere an satisty the recursion a
QUESTION 2 (Chapter 12, Exercises 787, p289) Let n be a positive integer. Let 1,1 E MEXa(R) satisfy the condition that Σ-laul 1 i < n. Show that Ic] < 1 for all c E spec(A). 1 for all
50. What is wrong with this "proof? "Theorem For every positive integer n = (n + /2. Basis Step: The formula is true for n = 1. Inductive Step: Suppose that +Y/2. Then -(+972 +*+- +*+1)/2 + + + /- + 1). By the inductive hypothesis, we have + /2-[(++P/2, completing the + inductive step.
Find the smallest positive integer n such that there are non-isomorphic simple graphs on n vertices that have the same chromatic polynomial. Explain carefully why the n you give as your answer is indeed the smallest.
Denote that /2 sin(2) 1 dar, 0 and for a positive integer N, denote that, N In = sin( () 2N пј j=1 Please provide an estimate for \I – IN), then please find a better approximation for I, so that the error can be controlled by Ns for some constant C, which is independent of N.