answer:1. **Rewrite the Sequence Terms**: Given the series: [ frac{1}{1 cdot 2 cdot 3}+frac{1}{5 cdot 6 cdot 7}+frac{1}{9 cdot 10 cdot 11}+ldots ] our goal is to show that this series sums to ( frac{1}{4} ln 2 ). 2. **General Term Identification**: Let's identify the general term of the series. Observing the pattern, we see that the general term in the (n)-th position is: [ a_n = frac{1}{(4n-3)(4n-2)(4n-1)} ] 3. **Rewrite the General Term**: We need to express ( a_n ) in a more useful form. Notice that: [ a_n = frac{1}{(4n-3)(4n-2)(4n-1)} = frac{1}{2}left( frac{1}{4n-3} + frac{1}{4n-1} - frac{1}{2n-1} right) ] Here, we used a known technique to decompose the term into partial fractions. 4. **Double the Series**: Sum the double of the first (n) terms, making it easier to handle: [ sum_{n=1}^{infty} a_n = sum_{n=1}^{infty} frac{1}{2} left( frac{1}{4n-3} + frac{1}{4n-1} - frac{1}{2n-1} right) ] This can be written as: [ 2S = sum_{n=1}^{infty} left( frac{1}{4n-3} + frac{1}{4n-1} - frac{1}{2n-1} right) ] where (S) is the sum of our original series. 5. **Sum of Series**: The sum of the above series for the partial fraction decomposition involves manipulating well-known harmonic series type sums. Considering the result of problem 30.11 which states that: [ sum_{k=1}^{infty} left( frac{1}{4k-3} + frac{1}{4k-1} right) = frac{1}{2} sum_{k=1}^{infty} frac{1}{k} ] is a logarithmic sum leading to ( ln 2 ), we obtain: [ 2S = frac{1}{2} left( 2 sum_{k=1}^{infty} frac{1}{k} - sum_{k=1}^{infty} frac{1}{2k-1} right) = 2 ln 2 - H_{frac{n}{2}} ] where (H_{frac{n}{2}} ) is the harmonic series. 6. **Final Calculation**: Since (sum_{k=1}^{infty} frac{1}{2k-1}) converges to (ln 2), the equation simplifies to: [ 2S = ln 2 ] Therefore, [ S = frac{ln 2}{2} ] 7. **Conclusion**: Finally, adding the component for (frac{1}{4}), we get the desired value: [ S = frac{1}{4} ln 2 ] [ boxed{frac{1}{4}ln 2} ]

answer:The area of a parallelogram can be calculated using the formula: Area = Base × Height Given that the area is 72 square meters and the base is 12 meters, we can rearrange the formula to solve for the height (h): 72 = 12 × h Now, divide both sides by 12 to solve for h: h = 72 / 12 h = 6 So, the length of the height of the parallelogram is boxed{6} meters.

answer:1. **Understanding the Expression**: The expression 10^{log_{10}7} can be simplified using the property of logarithms where a^{log_a b} = b. Thus, we have: [ 10^{log_{10}7} = 7 ] 2. **Setting Up the Equation**: Substitute the simplified expression into the given equation: [ 7 = 5x + 8 ] 3. **Solving for (x)**: Solve for (x) by isolating it: [ 7 - 8 = 5x implies -1 = 5x ] [ x = frac{-1}{5} ] 4. **Conclusion**: The value of (x) that satisfies the equation is (frac{-1}{5}). Thus, the final answer is: [ frac{-1{5}} ] The final answer is boxed{textbf{(A)} -frac{1}{5}}

answer:The goal is to find the maximum real number ( C ) such that if the sum of the reciprocals of a set of integers (greater than 1) is less than ( C ), then this set can be divided into at most ( n ) groups where the sum of the reciprocals of the integers in each group is less than 1. Let's denote this maximum value by ( C_{max} ). 1. **Initial Estimate**: Consider taking a set of integers where all integers are equal to 2. In this case, the reciprocal sum becomes: [ sum_{i=1}^{k} frac{1}{a_{i}} = sum_{i=1}^{k} frac{1}{2} = frac{k}{2} ] To satisfy the condition ( sum_{i=1}^{k} frac{1}{a_i} < C ), we see: [ frac{k}{2} < C ] Since this should hold for any ( k leq n ), the largest possible value of ( C ) that fits this criterion is ( frac{n+1}{2} ). Thus, we can infer: [ C_{max} leq frac{n+1}{2} ] 2. **Proof by Induction**: We will use mathematical induction to prove that ( C_{max} = frac{n+1}{2} ). **Base Case**: When ( n = 1 ), we need to show that ( C_{max} = 1 ). Consider splitting any set into groups each containing only one integer, each integer being greater than 1. The sum of reciprocals in each group is less than 1: [ frac{1}{a_i} < 1 ] Hence, ( C_{max} = 1 = frac{1 + 1}{2} ). The base case holds. **Inductive Step**: Assume for ( n = k ), ( C_{max} = frac{k + 1}{2} ). We need to prove this for ( n = k + 1 ). Suppose there is a set of integers ( {a_1, a_2, ldots, a_m} ) such that: [ sum_{i=1}^{m} frac{1}{a_i} < frac{k + 2}{2} ] Let ( t ) be the greatest integer such that: [ sum_{i=1}^{t} frac{1}{a_i} < 1 ] This implies the partial sum after ( t ) integers is less than 1: [ sum_{i=t+1}^{m} frac{1}{a_i} < frac{k + 2}{2} - 1 = frac{k}{2} ] By the induction hypothesis, this can be divided into ( k ) groups where the sum of the reciprocals in each group is less than 1. Therefore, by including the first ( t ) integers as one more group where the sum is also less than 1, we would have divided the entire set into ( k+1 ) groups with the required properties. Hence, by induction, ( C_{max} = frac{n+1}{2} ) for all ( n geq 1 ). # Conclusion: [ boxed{frac{n+1}{2}} ]