answer:First, evaluate the operation a diamond b. Given by the definition: [ a diamond b = frac{3a}{b} cdot frac{b}{a} = frac{3ab}{ab} = 3. ] Thus, regardless of what a and b are (as long as neither are zero), a diamond b will always be 3. Next, evaluate (7 diamond (4 diamond 9)) diamond 2. Starting with the innermost operation: [ 4 diamond 9 = 3. ] Then: [ 7 diamond 3 = 3. ] Finally: [ 3 diamond 2 = 3. ] Thus, the entire expression simplifies to boxed{3}.

answer:# Solution: Part (1): Finding the General Formulas 1. **Given Information:** - a_{1}=2b_{1}=2 - a_{2}=2b_{2} - a_{3}=2b_{3}+2 2. **Common Ratio and Difference:** Let the common ratio of {a_{n}} be q and the common difference of {b_{n}} be d. 3. **From a_{1}=2b_{1}=2:** - a_{1}=2 - b_{1}=1 4. **From a_{2}=2b_{2}:** - 2q=2(d+1) - Simplifying gives q=d+1 5. **From a_{3}=2b_{3}+2:** - 2q^{2}=2(1+2d)+2 - Simplifying gives q^{2}=2d+2 - Further simplification gives q^{2}=2(d+1) 6. **Solving for q:** - q^{2}=2q - Solving gives q=2 or q=0 (discard q=0) 7. **Therefore:** - q=2 - d=1 8. **General Formulas:** - a_{n}=2^{n} - b_{n}=n boxed{a_{n}=2^{n}, b_{n}=n} Part (2): Proof that {T}_{n}≤frac{9}{16} 1. **Given:** - c_{n}=frac{b_{n}^{2}}{a_{n}}=frac{n^{2}}{2^{n}}, for nin N^{*} 2. **Difference between Consecutive Terms:** - c_{n+1}-c_{n}=frac{(n+1)^{2}}{2^{n+1}}-frac{n^{2}}{2^{n}}=frac{-n^{2}+2n+1}{2^{n+1}} 3. **Behavior of c_{n}:** - For n=1 and 2, c_{n+1}-c_{n} > 0 - For ngeqslant 3, c_{n+1}-c_{n} < 0 - This implies c_{1} < c_{2} < c_{3} > c_{4} > c_{5} > cdots 4. **Values of c_{n} for Specific n:** - c_{1}=frac{1}{2}, c_{2}=1, c_{3}=frac{9}{8}, c_{4}=1, c_{5}=frac{25}{32} 5. **Product of First n Terms, T_{n}:** - For n=1 and 2, T_{n}=frac{1}{2} - For n=3 and 4, T_{n}=frac{9}{16} - For ngeqslant 5, since c_{n} < 1, we have T_{n+1} < T_{n} 6. **Conclusion:** - Hence, for all n, T_{n} leq frac{9}{16} boxed{T_{n} leq frac{9}{16}}

answer:To solve for B-N+T-Q, we start by breaking the components into their real and imaginary parts: - Real part of B-N+T-Q: (6-3-5)+(0-0-0-0)=6-3-5=6-8=-2 - Imaginary part of B-N+T-Q: (-5-2-2)+(0+0+0-0)=-5-2-2=-9 Thus, combining the real and imaginary components, the result is: B-N+T-Q = -2 - 9i Therefore, the answer is boxed{-2-9i}.

answer:We are given a function ( f: mathbb{Z}^2 to mathbb{R}^+ ) such that for all ((m, n) in mathbb{Z}^2), the value of ( f(m, n) ) is the arithmetic mean of the values at the four adjacent points. Formally, this can be written as: [ f(m, n) = frac{f(m+1, n) + f(m-1, n) + f(m, n+1) + f(m, n-1)}{4} ] We need to prove that ( f ) is constant, i.e., ( f(m, n) = c ) for some constant ( c ) and for all ((m, n) in mathbb{Z}^2). 1. **Define the Unbiased Random Walk:** Consider an unbiased random walk ({X_n}_{n in mathbb{N}}) on (mathbb{Z}^2) starting at ((0, 0)). This means that at each step, the walk moves to one of the four adjacent points with equal probability ( frac{1}{4} ). 2. **Define the Martingale:** Let ( Y_n = f(X_n) ). Since ( f ) satisfies the mean value property, ( {Y_n}_{n in mathbb{N}} ) is a martingale. This is because: [ mathbb{E}[Y_{n+1} mid Y_1, Y_2, ldots, Y_n] = mathbb{E}[f(X_{n+1}) mid X_n] = f(X_n) = Y_n ] Here, the expectation is taken over the four possible moves of the random walk. 3. **Apply Doob's Forward Convergence Theorem:** By Doob's Forward Convergence Theorem, a non-negative martingale converges almost surely. Therefore, there exists a constant ( c ) such that: [ lim_{n to infty} Y_n = c quad text{almost surely} ] 4. **Implication for the Function ( f ):** The convergence ( lim_{n to infty} Y_n = c ) implies that ( f ) can be made arbitrarily close to ( c ) along 'most' of the boundary of any sufficiently large square. Since ( f ) at the center of the square is a weighted average of ( f ) along the boundary, ( f ) must be arbitrarily close to ( c ) at the center of the square as well. 5. **Conclusion:** Since the argument holds for any point ((m, n) in mathbb{Z}^2), it follows that ( f(m, n) = c ) for all ((m, n) in mathbb{Z}^2). Thus, ( f ) is constant. (blacksquare)