answer:1. From 6S_n=a_n^2+3a_n+2 (equation 1), we get 6S_{n-1}=a_{n-1}^2+3a_{n-1}+2 (equation 2). Subtract equation 2 from equation 1, we get (a_n+a_{n-1})(a_n-a_{n-1}-3)=0. Since all terms in the sequence {a_n} are positive, we have a_n-a_{n-1}=3. Therefore, the sequence {a_n} is an arithmetic sequence with the first term 1 and common difference 3. Hence, the general term formula for the sequence {a_n} is a_n=3n-2. 2. From the sum of an arithmetic series, S_n=frac{3n^2-n}{2}. Thus, b_n=frac{2S_n}{3n-1}cdot 2^n=ncdot 2^n. Now we need to find the sum of the first n terms of the sequence {b_n}, denoted as T_n. T_n=1cdot 2^1+2cdot 2^2+dots+ncdot 2^n (equation 3), 2T_n=1cdot 2^2+2cdot 2^3+dots+ncdot 2^{n+1} (equation 4). Subtracting equation 3 from equation 4, we get -T_n=2^1+2^2+2^3+dots+2^n-ncdot 2^{n+1}=frac{2(1-2^n)}{1-2}-ncdot 2^{n+1}=(1-n)2^{n+1}-2. Hence, the sum of the first n terms of the sequence {b_n} is boxed{T_n=(n-1)2^{n+1}+2}.

answer:Let's denote the total number of marbles in the bag as ( T ), and the number of white marbles as ( W ). We know that the probability of selecting a red or white marble is 0.9, and we have the number of red marbles, which is 9. So, the probability of selecting a red or white marble can be expressed as: [ frac{9 + W}{T} = 0.9 ] We also know that the total number of marbles is the sum of blue, red, and white marbles: [ T = 5 + 9 + W ] [ T = 14 + W ] Now, we can substitute ( T ) from the second equation into the first equation: [ frac{9 + W}{14 + W} = 0.9 ] Now, we solve for ( W ): [ 9 + W = 0.9(14 + W) ] [ 9 + W = 12.6 + 0.9W ] [ W - 0.9W = 12.6 - 9 ] [ 0.1W = 3.6 ] [ W = frac{3.6}{0.1} ] [ W = 36 ] So, there are 36 white marbles in the bag. Now, we can find the total number of marbles ( T ): [ T = 14 + W ] [ T = 14 + 36 ] [ T = 50 ] Therefore, there are boxed{50} marbles in the bag.

answer:1. **Define the sequences and initial conditions:** We are given the sequences (a_n) and (b_n) with initial conditions: [ a_0 = 1, quad a_1 = 4, quad a_2 = 49 ] and the recurrence relations: [ begin{cases} a_{n+1} = 7a_n + 6b_n - 3, b_{n+1} = 8a_n + 7b_n - 4. end{cases} ] 2. **Express (a_{n+1}) in terms of previous terms:** We start by expressing (a_{n+1}) in terms of (a_n) and (b_n): [ a_{n+1} = 7a_n + 6b_n - 3 ] We also have: [ 7a_n = 49a_{n-1} + 42b_{n-1} - 21 ] Subtracting the second equation from the first: [ a_{n+1} - 7a_n = (7a_n + 6b_n - 3) - (49a_{n-1} + 42b_{n-1} - 21) ] Simplifying, we get: [ a_{n+1} - 7a_n = 6b_n - 3 - 49a_{n-1} - 42b_{n-1} + 21 ] 3. **Use the recurrence relation for (b_n):** We know: [ b_n - 7b_{n-1} = 8a_{n-1} - 4 ] Substituting this into the previous equation: [ a_{n+1} - 7a_n = 6(7b_{n-1} + 8a_{n-1} - 4) - 3 - 49a_{n-1} - 42b_{n-1} + 21 ] Simplifying further: [ a_{n+1} - 7a_n = 42b_{n-1} + 48a_{n-1} - 24 - 3 - 49a_{n-1} - 42b_{n-1} + 21 ] [ a_{n+1} - 7a_n = -a_{n-1} - 6 ] Rearranging, we get: [ a_{n+1} = 14a_n - a_{n-1} - 6 ] 4. **Form the characteristic equation:** The recurrence relation is: [ a_{n+1} - 15a_n + 15a_{n-1} - a_{n-2} = 0 ] The characteristic equation is: [ x^3 - 15x^2 + 15x - 1 = 0 ] 5. **Solve the characteristic equation:** One root is obviously (x = 1). The other roots can be found by factoring or using the quadratic formula: [ (x - 1)(x^2 - 14x + 1) = 0 ] Solving the quadratic equation: [ x = frac{14 pm sqrt{196 - 4}}{2} = 7 pm 4sqrt{3} ] So the roots are (1, 7 + 4sqrt{3}, 7 - 4sqrt{3}). 6. **Form the general solution:** The general solution to the recurrence relation is: [ a_n = A cdot 1^n + B cdot (7 + 4sqrt{3})^n + C cdot (7 - 4sqrt{3})^n ] Using the initial conditions (a_0 = 1), (a_1 = 4), and (a_2 = 49), we solve for (A), (B), and (C): [ a_n = frac{1}{2} + frac{(7 + 4sqrt{3})^n}{4} + frac{(7 - 4sqrt{3})^n}{4} ] Simplifying, we get: [ a_n = left( (2 + sqrt{3})^n + (2 - sqrt{3})^n right)^2 cdot frac{1}{4} ] 7. **Prove (a_n) is a perfect square:** Notice that: [ a_n = left( (2 + sqrt{3})^n + (2 - sqrt{3})^n right)^2 ] Since ((2 + sqrt{3})^n + (2 - sqrt{3})^n) is an integer for all (n), (a_n) is a perfect square. (blacksquare)

answer:1. **Define the Problem and Given Conditions:** We need to prove that a subset ( S ) of ( mathbb{N} ) that is closed under addition is eventually linear. This means there exist ( k, N in mathbb{N} ) such that for ( n > N ), ( n in S ) if and only if ( k mid n ). 2. **Characterize the Submonoids:** The problem can be rephrased as characterizing all submonoids of the monoid ( mathbb{N}_0 ). A submonoid of ( mathbb{N}_0 ) is a subset that is closed under addition and contains the identity element (0). 3. **Initial Elements and Their GCD:** Let ( a_1, a_2 in S ). Since ( S ) is closed under addition, any linear combination ( m = x a_1 + y a_2 ) with ( x, y geq 0 ) is also in ( S ). Consider the greatest common divisor ( c_1 = gcd(a_1, a_2) ). For sufficiently large ( m ), ( m ) can be expressed as ( m = x a_1 + y a_2 ), implying ( m ) is a multiple of ( c_1 ). 4. **Extending to More Elements:** If there exists another element ( a_3 in S ) that is not a multiple of ( gcd(a_1, a_2) ), we extend our consideration to ( gcd(a_1, a_2, a_3) ). We repeat this process iteratively. 5. **Finite Termination:** This process must terminate because the number of different residues modulo ( gcd(a_1, a_2, a_3, ldots, a_i) ) strictly decreases with each step. Eventually, we reach a point where all elements of ( S ) are multiples of some ( k ), and for sufficiently large ( n ), ( n in S ) if and only if ( k mid n ). 6. **Conclusion:** Therefore, ( S ) is eventually linear, as there exist ( k, N in mathbb{N} ) such that for ( n > N ), ( n in S ) if and only if ( k mid n ). (blacksquare)