answer:Starting with the sequence { a_n } and the recurrence 2a_{n+1} = a_n + a_{n+2}, we deduce that { a_n } must be an arithmetic sequence since it follows a linear recurrence relation. With a_1 = 2 and a_2 = 4, the common difference, d, is equal to 2. Therefore, the general term is a_n = 2 + 2(n - 1) = 2n. For the sequence { b_n } and the inequality b_{n+1} - b_n < 2^n + frac{1}{2}, we infer b_{n+2} - b_{n+1} < 2^{n+1} + frac{1}{2}, which further implies that b_{n+2} - b_n < 3 cdot 2^n + frac{1}{2}. However, since we also have b_{n+2} - b_n > 3 cdot 2^n - 1 and b_n in mathbb{Z}, it follows that b_{n+2} - b_n = 3 cdot 2^n. Using initial conditions b_1 = 2 and b_2 = 4, we can now derive the general term for b_n. When n = 2k - 1 (k geq 2), we write: b_n = (b_n - b_{n-2}) + (b_{n-2} - b_{n-4}) + cdots + (b_3 - b_1) + b_1 b_n = 3 cdot (2^{n-2} + 2^{n-4} + cdots + 2^3 + 2) + 2 b_n = 3 cdot frac{2(4^{k-1} - 1)}{4 - 1} + 2 b_n = 2^{2k-1} = 2^n , which holds true for n = 1 as well. For even n = 2k (k geq 1), we have: b_n = (b_n - b_{n-2}) + (b_{n-2} - b_{n-4}) + cdots + (b_4 - b_2) + b_2 b_n = 3 cdot (2^{n-2} + 2^{n-4} + cdots + 2^4 + 4) + 4 b_n = 3 cdot frac{4(4^{k-1} - 1)}{4 - 1} + 4 b_n = 4^k = 2^n . From these two cases, we conclude that b_n = 2^n for all n. Consequently, frac{nb_n}{a_n} = frac{n cdot 2^n}{2n} = 2^{n - 1}. The sum of the first n terms of the sequence left{ frac{nb_n}{a_n} right} is therefore: sum_{k=1}^n 2^{k-1} = frac{2^n - 1}{2 - 1} = 2^n - 1 . So, the answer to the problem is boxed{2^n - 1}.

answer:1. **Initialize the Setup:** We consider ( n ) lines on a plane such that no two lines are parallel, and no three lines are concurrent. We need to prove that each of these ( n ) lines creates a triangular region with its neighboring intersections. 2. **Consider the Line ( l ):** Choose any particular line ( l ). The intersections of all other ( n - 1 ) lines with ( l ) will divide ( l ) into ( n ) segments. 3. **Observation of Intersections:** The other ( n - 1 ) lines will intersect among themselves and with the line ( l ) at points. Let these intersections create ( A ) regions. According to the problem conditions, lines are neither parallel nor concurrent, so there will be ( binom{n}{2} ) intersections in total. 4. **Count the Regions:** The ( n ) lines will divide the plane into ( frac{n(n-1)}{2} + n + 1 ) regions. However, we are interested specifically in regions near the chosen line ( l ). 5. **Focus on ( l ) and its Adjacent Regions:** Each finite segment created on ( l ) by other lines will have finite adjacent regions. To understand the nature of these regions: - Consider ( l_1 ) and ( l_2 ) as the nearest lines intersecting ( l ). - Each pair ( (l, l_i, l_j) ) where ( l_i ) and ( l_j ) are any two lines other than ( l ), will form vertex triangles at each intersection point. 6. **Finite Triangular Region Adjacent to ( l ):** Since no three lines concur, the intersection points along ( l ) uniquely determine triangular regions. The adjacent finite portion of the plane will always carve out a triangular section because: - Each segment on ( l ) determined by intersections with ( l_i ) and ( l_j ) will form triangular regions due to the finite and non-parallel intersection properties. 7. **Conclusion from ( A ):** Among the created regions ( l_1 times l_2 = A ) nearest to the considered line ( l ), there necessarily exists a triangular region. Conclusively, **each line among the ( n ) will always have a neighboring region that is triangular.** Therefore, the proof is accomplished as this applies universally to any line ( l ). [ boxed{} ]

answer:1. Given the problem in triangle ( triangle ABC ), the angle ratio is ( A:B:C = 4:2:1 ). 2. Using the fact from the Law of Sines, we can express the sides in terms of the sine of the angles: [ a = 2R sin(A), quad b = 2R sin(B), quad c = 2R sin(C) ] where ( R ) is the circumradius of the triangle. 3. Since the angles have a ratio of ( 4:2:1 ), we can express them as: [ A = 4x, quad B = 2x, quad C = x ] where ( x ) is some angular measure. 4. Since ( A + B + C = pi ) (sum of angles in a triangle): [ 4x + 2x + x = 7x = pi implies x = frac{pi}{7} ] Therefore: [ A = 4 left(frac{pi}{7}right) = frac{4pi}{7}, quad B = 2 left(frac{pi}{7}right) = frac{2pi}{7}, quad C = frac{pi}{7} ] 5. Applying these angle measures we have: [ a = 2R sin left(frac{4pi}{7}right), quad b = 2R sin left(frac{2pi}{7}right), quad c = 2R sin left(frac{pi}{7}right) ] 6. We need to prove the identity: [ frac{1}{a} + frac{1}{b} = frac{1}{c} ] 7. Substitute the expressions for (a), (b), and (c) in terms of ( R ): [ frac{1}{a} = frac{1}{2R sin left(frac{4pi}{7}right)} = frac{1}{2R} cdot frac{1}{sin left(frac{4pi}{7}right)} ] [ frac{1}{b} = frac{1}{2R sin left(frac{2pi}{7}right)} = frac{1}{2R} cdot frac{1}{sin left(frac{2pi}{7}right)} ] [ frac{1}{c} = frac{1}{2R sin left(frac{pi}{7}right)} = frac{1}{2R} cdot frac{1}{sin left(frac{pi}{7}right)} ] 8. Adding the fractions: [ frac{1}{a} + frac{1}{b} = frac{1}{2R} left[frac{1}{sin left(frac{4pi}{7}right)} + frac{1}{sin left(frac{2pi}{7}right)} right] ] 9. Now using the trigonometric identity: [ sin X + sin Y = 2 sin left(frac{X+Y}{2} right) cos left(frac{X-Y}{2} right) ] for ( X = frac{4pi}{7} ) and ( Y = frac{2pi}{7} ): [ sin left(frac{4pi}{7}right) + sin left(frac{2pi}{7}right) = 2 sin left(frac{4pi/7 + 2pi/7}{2} right) cos left(frac{4pi/7 - 2pi/7}{2} right) ] [ = 2 sin left(frac{6pi}{7} cdot frac{1}{2} right) cos left(frac{2pi}{7} cdot frac{1}{2} right) ] [ = 2 sin left(frac{3pi}{7} right) cos left(frac{pi}{7} right) ] 10. Simplifying the right-hand side of the fraction: [ frac{1}{sin left(frac{4pi}{7}right) sin left(frac{2pi}{7}right)} = frac{1}{sin left(frac{pi}{7}right)} ] 11. Substitution gives: [ frac{2 sin left(frac{3pi}{7} right) cos left(frac{pi}{7} right)}{2R sin left(frac{3pi}{7}right) cos left(frac{pi}{7}right)} = frac{1}{sin left(frac{pi}{7}right)} = frac{1}{c} ] # Conclusion: [ boxed{frac{1}{a} + frac{1}{b} = frac{1}{c}} ]

answer:The projection of mathbf{v} onto mathbf{w} is given by the formula: [text{proj}_{mathbf{w}} mathbf{v} = frac{mathbf{v} cdot mathbf{w}}{mathbf{w} cdot mathbf{w}} mathbf{w}] Calculating the dot products: [mathbf{v} cdot mathbf{w} = 3 times 2 + 4 times (-1) = 6 - 4 = 2] [mathbf{w} cdot mathbf{w} = 2 times 2 + (-1) times (-1) = 4 + 1 = 5] Thus, [text{proj}_{mathbf{w}} mathbf{v} = frac{2}{5} begin{pmatrix} 2 -1 end{pmatrix} = begin{pmatrix} frac{4}{5} -frac{2}{5} end{pmatrix}] Conclusion: [boxed{begin{pmatrix} frac{4}{5} -frac{2}{5} end{pmatrix}}]