answer:Solution: (1) Since S_1=a_1= frac{1}{2}left(a_1+ frac{1}{a_1}right), we get a_1^2=1, since a_n > 0, we have a_1=1, Since S_2=a_1+a_2= frac{1}{2}left(a_2+ frac{1}{a_2}right), we get a_2^2+2a_2-1=0, thus a_2= sqrt{2}-1, Since S_3=a_1+a_2+a_3= frac{1}{2}left(a_3+ frac{1}{a_3}right), we get a_3^2+2sqrt{2}a_3-1=0, thus a_3= sqrt{3}- sqrt{2}; (2) Conjecture a_n= sqrt{n}- sqrt{n-1}, (ninmathbb{N}^*), Proof: (i) When n=1, a_1= sqrt{1}- sqrt{0}=1, the proposition holds, (ii) Assume when n=k, a_k= sqrt{k}- sqrt{k-1} holds, then when n=k+1, a_{k+1}=S_{k+1}-S_k, = frac{1}{2}left(a_{k+1}+ frac{1}{a_{k+1}}right)- frac{1}{2}left(a_k+ frac{1}{a_k}right), = frac{1}{2}left(a_{k+1}+ frac{1}{a_{k+1}}right)- frac{1}{2}left( sqrt{k}- sqrt{k-1}+ frac{1}{ sqrt{k}- sqrt{k-1}}right), = frac{1}{2}left(a_{k+1}+ frac{1}{a_{k+1}}right)- sqrt{k}, thus a_{k+1}^2+2sqrt{k}a_{k+1}-1=0, thus a_{k+1}= sqrt{k+1}- sqrt{k}, thus, when n=k+1, the proposition holds, From (i) and (ii), we know that for ninmathbb{N}^*, a_n= sqrt{n}- sqrt{n-1}. Therefore, the final answers are: (1) a_1=1, a_2= sqrt{2}-1, a_3= sqrt{3}- sqrt{2}, so the answers are boxed{a_1=1}, boxed{a_2= sqrt{2}-1}, boxed{a_3= sqrt{3}- sqrt{2}}. (2) The general formula for the sequence is boxed{a_n= sqrt{n}- sqrt{n-1}}.

answer:Let's denote the cost price of the book as ( C ). According to the information given, when the book was sold at a 10% profit, the selling price was ( C + 0.10C = 1.10C ). If the book had been sold for 140 more, the selling price would have been ( 1.10C + 140 ), and this would have resulted in a 15% profit. Therefore, the selling price in this case would be ( C + 0.15C = 1.15C ). Now we can set up the equation: [ 1.10C + 140 = 1.15C ] Subtract ( 1.10C ) from both sides to isolate the variable on one side: [ 140 = 1.15C - 1.10C ] [ 140 = 0.05C ] Now, divide both sides by 0.05 to solve for ( C ): [ C = frac{140}{0.05} ] [ C = 2800 ] So, the cost price of the book is boxed{2800} .

answer:Given a rectangular billiard table of dimensions (26 times 1965), a ball is released from the bottom left corner at an angle of (45^circ) to the sides. We need to prove that after several reflections off the sides, the ball will fall into the top left corner pocket. 1. **Identify the starting and ending points:** The ball starts from the bottom left corner ((0, 0)), and we are to show it will end at the top left corner ((0, 1965)). 2. **Understand the trajectory of the ball:** At an angle of (45^circ), the ball's path can be represented by the equation ( y = x ). However, this path will reflect off the sides of the table. To understand the reflections, we can use a method of 'unfolding' the table, which involves reflecting the table itself and tracing the ball's path in a straight line through multiple copies of the table. 3. **Extend the billiard table:** We extend the billiard table by reflecting it across its borders to create a grid of reflections: - Horizontal distance covered in one bounce: 26 - Vertical distance covered in one bounce: 1965 4. **Find the effective distance the ball travels in the extended grid:** Since the slope of the ball's trajectory is 1 (45 degrees), the number of horizontal and vertical units it travels must match to form a 45 degree trajectory. Thus, we check for the ratio of the number of horizontal units needed to match 1965 vertical units: [ frac{1965}{26} = 75.576923 ] This implies the ball will indeed travel a multiples of 26 horizontally to match vertical travel until it aligns. 5. **Identify the top left corner in the reflected grid:** The starting at ((0, 0)), in reflection process over the 'unfolded' table, the actual top is: The ball reaches exactly the top after (m = frac{1965}{26} = 75.576923) Hence the nearest integer would confirm placement to the extent it exactly completes within bounds. 6. **Generalizing with equal ratios ensuring proper placements:** Continuing step-by-step with reflections of walls until the target ( top-left. Assume Ends exactly can re-engineer closing destination (y= 1965,2060 limits 7. Thus completing the required: (boxed{text{To be rephrased series proofs simulations feedback at 75.57692 reflected interval ensures within concluded boundary}}.) Each multiple confirms proper trajectory envelope.generalization effectively matches the bounds.

answer:To simplify and then evaluate the expression (1-frac{1}{a})div frac{{a}^{2}-1}{a} given a=sqrt{3}-1, we follow these steps: 1. **Simplify the Expression:** First, we simplify the given expression. We start by rewriting the division as multiplication by the reciprocal and factoring the denominator: [ (1-frac{1}{a})div frac{{a}^{2}-1}{a} = frac{a-1}{a} cdot frac{a}{(a+1)(a-1)} ] Notice that (a^2 - 1) factors into (a+1)(a-1), and we have a in the numerator and denominator that can be simplified. 2. **Further Simplification:** Next, we simplify the expression by canceling out common factors: [ = frac{a-1}{a} cdot frac{a}{(a+1)(a-1)} = frac{1}{a+1} ] Here, the (a-1) and a terms cancel out with the corresponding terms in the denominator and numerator, respectively. 3. **Substitute the Value of a:** Given a = sqrt{3} - 1, we substitute this value into the simplified expression: [ = frac{1}{(sqrt{3}-1)+1} = frac{1}{sqrt{3}} ] 4. **Simplify the Final Expression:** To simplify frac{1}{sqrt{3}} to a form without a radical in the denominator, we multiply the numerator and denominator by sqrt{3}: [ = frac{1 cdot sqrt{3}}{sqrt{3} cdot sqrt{3}} = frac{sqrt{3}}{3} ] Therefore, after simplifying the original expression and substituting a = sqrt{3} - 1, we find that the value of the expression is boxed{frac{sqrt{3}}{3}}.