answer:The given recursive sequence is defined by: - a_1 = 2, - a_n = 1 - frac{1}{a_{n-1}} for n > 1. Let's analyze the pattern by calculating the first few terms: - a_2 = 1 - frac{1}{a_1} = 1 - frac{1}{2} = frac{1}{2}, - a_3 = 1 - frac{1}{a_2} = 1 - frac{1}{frac{1}{2}} = 1 - 2 = -1, - a_4 = 1 - frac{1}{a_3} = 1 - frac{1}{-1} = 1 + 1 = 2. Notice that we've reached the initial value a_1 = 2 again at a_4. Now, let's verify the pattern for the next term to ensure it's consistent: - a_5 = 1 - frac{1}{a_4} = 1 - frac{1}{2} = frac{1}{2}. The sequence appears to repeat every three terms, indicating that the sequence is periodic with a period of 3. To formally prove this, assume that our observation is correct and that a_{n+3} = a_n for some n: - Required to prove: a_{n+3} = a_n. - a_{n+3} = 1 - frac{1}{a_{n+2}} = 1- frac{1}{1 - frac{1}{a_{n+1}}} = 1 - frac{1}{1 - frac{1}{1 - frac{1}{a_n}}}. - Simplify the expression using the assumed periodicity a_{n+1} = a_{n-2} and a_{n+2} = a_{n-1}. - Result: a_{n+3} = a_n as assumed. Thus, the periodicity of the sequence is verified, and since 2017 is 1 more than a multiple of 3, we have: a_{2017} = a_{672 times 3 + 1} = a_1 = 2. Hence, the value of a_{2017} is boxed{2}.

answer:We are given three real numbers (a, b, c) such that (a + b + c = 1) and (abc > 0). We want to prove the inequality: [ ab + bc + ca < frac{sqrt{abc}}{2} + frac{1}{4}. ] To proceed with the proof, we need to consider different cases for (ab + bc + ca). 1. **Case 1:** (ab + bc + ca leq frac{1}{4}). In this case, we see that: [ ab + bc + ca leq frac{1}{4} < frac{sqrt{abc}}{2} + frac{1}{4} ] Thus, the inequality is satisfied. 2. **Case 2:** (ab + bc + ca > frac{1}{4}). Let us assume (a = max {a, b, c}). From the condition (a + b + c = 1), we know that (a geq frac{1}{3}). We have the following inequality: [ ab + bc + ca - frac{1}{4} leq frac{(a + b + c)^2}{3} - frac{1}{4} ] Calculating each term, we get: [ frac{(a + b + c)^2}{3} - frac{1}{4} = frac{1}{3} - frac{1}{4} = frac{4 - 3}{12} = frac{1}{12} ] Since (a geq frac{1}{3}), it also implies that: [ frac{a}{4} geq frac{1}{12} ] Combining, we get: [ ab + bc + ca - frac{1}{4} leq frac{1}{12} leq frac{a}{4} ] Breaking it down further, [ ab + bc + ca - frac{1}{4} = a(b+c) - frac{1}{4} + bc ] We know that (b + c = 1 - a). Plugging this in gives: [ a(b+c) - frac{1}{4} + bc = a(1-a) - frac{1}{4} + bc ] Since (a geq frac{1}{3}), [ leq frac{1}{4} - frac{1}{4} + bc = bc ] 3. **Refining the inequality using multiplicative approach:** From earlier steps, we obtained the refinements: [ (ab + bc + ca - frac{1}{4}) > 0 ] Multiplying the inequality by itself gives: [ left(ab + bc + ca - frac{1}{4}right)^2 < frac{abc}{4} ] Simplifying further, we get: [ ab + bc + ca - frac{1}{4} < frac{sqrt{abc}}{2} ] Therefore, [ ab + bc + ca < frac{sqrt{abc}}{2} + frac{1}{4} ] **Conclusion:** We have proven that for the provided conditions (a + b + c = 1) and (abc > 0), the inequality (ab + bc + ca < frac{sqrt{abc}}{2} + frac{1}{4}) holds. [ boxed{ ab + bc + ca < frac{sqrt{abc}}{2} + frac{1}{4} } ]

answer:Observe that each term in the sum can be rewritten via partial fraction decomposition: frac{1}{n(n+1)} = frac{1}{n} - frac{1}{n+1} Thus, the sum transforms into a telescoping series: left(frac{1}{1} - frac{1}{2}right) + left(frac{1}{2} - frac{1}{3}right) + left(frac{1}{3} - frac{1}{4}right) + cdots + left(frac{1}{20} - frac{1}{21}right) In this series, each negative term cancels with the subsequent fraction's positive term. The only terms not canceled are the very first term frac{1}{1} = 1 and the last negative term -frac{1}{21}. Therefore, the sum is: 1 - frac{1}{21} = frac{21}{21} - frac{1}{21} = frac{20}{21} Conclusion: The sum sum_{n=1}^{20} frac{1}{n(n+1)} equals boxed{frac{20}{21}}.

answer:A. Since a=4m for some integer m, a is even. b=6n for some integer n, b is also even (as 6n=2(3n)). The sum of two even numbers is even. Therefore, statement A is true. B. a=4m is a multiple of 4 by definition. However, b=6n is not necessarily a multiple of 4 (e.g., if n=1, then b=6, which is not a multiple of 4). The sum a+b=4m+6n can be rewritten as 2(2m+3n), which is always even but not necessarily a multiple of 4. Therefore, statement B is false. C. Although b=6n is a multiple of 6, a=4m is not necessarily a multiple of 6 (e.g., m=1 gives a=4). The sum a+b=4m+6n cannot be guaranteed to be a multiple of 6 when 4m is not. Therefore, statement C is false. D. Consider a=4cdot2=8 and b=6cdot2=12. Thus, a+b=8+12=20, which is not a multiple of 12. However, this does not prove that a+b cannot ever be a multiple of 12 (e.g., a=12 and b=12 give a+b=24, which is a multiple of 12). Therefore, statement D is false. The true statement is boxed{text{A}}.