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question:Let the sum of the first n terms of a geometric sequence {a_n} be S_n. If S_3=7 and S_6=63, find the value of a_7+a_8+a_9.
answer:By the properties of geometric series, S_3, S_6 - S_3, S_9 - S_6, ... form a new geometric sequence. Let's denote the common ratio of this new sequence as q.Given that S_3=7 and S_6=63, we have S_6 - S_3 = 63 - 7 = 56. Hence, the common ratio q of the series can be calculated as follows:q = frac{S_6 - S_3}{S_3} = frac{56}{7} = 8To find a_7 + a_8 + a_9, we need S_9 - S_6. As we already know the common ratio q, we can use it to find S_9 - S_6:S_9 - S_6 = (S_6 - S_3) cdot q = 56 cdot 8 = 448Therefore, the sum we are looking for is:a_7 + a_8 + a_9 = S_9 - S_6 = boxed{448}
question:1. [4 points] Around a flower, in the same plane as it, a bumblebee and a bee are flying along two circles. The speed of the bee is one and a half times the speed of the bumblebee. In the specified plane, a rectangular coordinate system is introduced, in which the flower (the common center of the circles) is at the point O(0 ; 0). The bee moves clockwise, while the bumblebee moves counterclockwise. At the initial moment, the bee and the bumblebee are at points M_{0}left(-frac{1}{2} ; frac{sqrt{3}}{2}right) and N_{0}(2 ; 0), respectively. Determine the coordinates of all positions of the bumblebee where the distance between it and the bee will be the shortest.
answer:Answer: (sqrt{3} ; 1),(-1 ; sqrt{3}),(-sqrt{3} ;-1),(1 ;-sqrt{3}).Solution. Let the points where the bee and the bumblebee are located be M(alpha) and N(beta), respectively, where alpha and beta are the angles that the radius vectors of points M and N form with the positive direction of the x-axis. Note that the angle between overrightarrow{A M_{0}} and overrightarrow{A N_{0}} is frac{2 pi}{3}, and at this point alpha_{0}=frac{2 pi}{3}, beta_{0}=0 - are the angles corresponding to the initial positions of the insects.The distance between the bumblebee and the bee will be the smallest when the angle between the vectors overrightarrow{A M} and overrightarrow{A N} is zero. Since left|A M_{0}right|=1 and left|A N_{0}right|=2 are the radii of the circles, and left|A N_{0}right|=2left|A M_{0}right|, the angular velocity of the bee is three times the angular velocity of the bumblebee.Let at the moment of the vectors overrightarrow{A M} and overrightarrow{A N} aligning, the bumblebee has moved by an angle omega. Then alpha_{0}-3 omega+2 pi k=beta_{0}+omega, where k in mathbb{Z}. Therefore, omega=frac{alpha_{0}-beta_{0}}{4}+frac{pi k}{2}=frac{pi}{6}+frac{pi k}{2}, where k=0,1,2 ldots There will be four different points (for k=0,1,2,3). For k=0, we get beta_{1}=beta_{0}+frac{pi}{6}=frac{pi}{6}. The coordinates of the bumblebee's position are found using the formulas x_{N}=2 cos beta_{1}=sqrt{3}, y_{N}=2 sin beta_{1}=1.The other points are obtained by rotating point N_{1} around the origin by angles frac{pi}{2}, pi, frac{3 pi}{2} and have, respectively, coordinates (-1 ; sqrt{3}),(-sqrt{3} ;-1),(1 ;-sqrt{3}).
question:A technique widely used to calculate summations is the Telescoping Sum. It consists of "decomposing" the terms of a sum into parts that cancel each other out. For example,begin{aligned}& frac{1}{1 cdot 2}+frac{1}{2 cdot 3}+frac{1}{3 cdot 4}+frac{1}{4 cdot 5}= & left(frac{1}{1}-frac{1}{2}right)+left(frac{1}{2}-frac{1}{3}right)+left(frac{1}{3}-frac{1}{4}right)+left(frac{1}{4}-frac{1}{5}right)= & frac{1}{1}-frac{1}{5}= & frac{4}{5}end{aligned}With this technique, we can find a way to sum consecutive odd numbers. Let's see:a) Counting the odd numbers one by one starting from 1, verify that the number at position m is equal to m^{2}-(m-1)^{2}.b) Calculate the sum of all odd numbers between 1000 and 2014.
answer:Solutiona) See that the first odd number is 2 cdot 1-1 and, knowing that odd numbers increase by 2, we can conclude that the odd number at position m in our count isbegin{aligned}2 cdot 1-1+underbrace{2+2+ldots+2}_{m-1 text { times }} & =2 cdot 1-1+2(m-1) & =2 m-1end{aligned}To verify that it matches the number in item a), we just need to calculatem^{2}-(m-1)^{2}=m^{2}-left(m^{2}-2 m-1right)=2 m-1b) We want to sum the odd numbers from 1001=2 cdot 501-1 to 2013=2 cdot 1007-1. Using the expression from item a), we havebegin{aligned}1001 & =501^{2}-500^{2} 1003 & =502^{2}-501^{2} 1005 & =503^{2}-502^{2} & cdots 2011 & =1006^{2}-1005^{2} 2013 & =1007^{2}-1006^{2}end{aligned}Adding everything, we see that all numbers from 501^{2} to 1006^{2} are canceled. Thus, the result is:begin{aligned}1001+1003+ldots+2013 & =1007^{2}-500^{2} & =(1007-500)(1007+500) & =507 cdot 1507 & =764049end{aligned}So the sum of the odd numbers between 1000 and 2014 is 763048.
question:If the function f(2x+1)=x^2-2x, then find f(3)=underline{qquadqquad}.
answer:Step 1: Understanding the ProblemThis problem requires knowledge of basic function concepts. We can solve for f(3) by setting 2x+1 equal to 3.Step 2: Solving for xLet's set up the equation:2x+1=3Subtracting 1 from both sides, we have:2x=2Finally, dividing by 2 gives us:x=1Step 3: Finding f(3)Now that we have x=1, substitute this value back into the original function:f(2x+1)=x^2-2xf(3)=(1)^2-2(1)f(3)=1-2f(3)=-1Therefore, the answer is boxed{-1}.
question:The reciprocal of -1 is ______.
answer:To find the reciprocal of a number, we look for a value that, when multiplied by the original number, gives us 1. For the number -1, we can set up the equation:[-1 times x = 1]We know that multiplying two negative numbers results in a positive number. Specifically, if we have -1 times -1, it equals 1. Thus, in our equation, x must be -1 to satisfy the condition -1 times x = 1. Therefore, the reciprocal of -1 is -1.[boxed{-1}]
question:The value of cos^2 15^circ - sin^2 15^circ is ( ).A: frac{1}{2}B: frac{sqrt{2}}{2}C: frac{sqrt{3}}{2}D: frac{sqrt{6}}{2}
answer:Solution: cos^2 15^circ - sin^2 15^circ = cos 2 times 15^circ = cos 30^circ = frac{sqrt{3}}{2}. Therefore, the correct choice is C.By simplifying the given expression using the double angle formula for cosine, and then using the trigonometric values of special angles, we can find the value. This problem tests the knowledge of the double angle formula for cosine, as well as the trigonometric values of special angles. Mastering the double angle formula for cosine is key to solving this problem.Thus, the answer is boxed{C}.