💡Okay, so I have this double sum to compute: the sum from n equals 2 to infinity of the sum from k equals 1 to n minus 1 of k over 3 to the power of n plus k. Hmm, that sounds a bit complicated, but maybe I can break it down.First, let me write it out to visualize it better:[sum_{n=2}^infty sum_{k=1}^{n-1} frac{k}{3^{n+k}}]Alright, so for each n starting from 2, I'm summing over k from 1 up to n-1. Each term is k divided by 3 raised to the sum of n and k. Maybe I can switch the order of summation to make it easier. Instead of summing over n first and then k, maybe I can sum over k first and then n. Let me try that.If I switch the order, the limits of summation will change. For each k starting from 1, n will go from k+1 to infinity because originally, k goes up to n-1, so n has to be at least k+1. So, rewriting the sum:[sum_{k=1}^infty sum_{n=k+1}^infty frac{k}{3^{n+k}}]Okay, that seems manageable. Now, looking at the inner sum, which is over n from k+1 to infinity. Let me factor out terms that don't depend on n. The term k is independent of n, so I can factor that out:[sum_{k=1}^infty k cdot sum_{n=k+1}^infty frac{1}{3^{n+k}}]Wait, actually, 3^{n+k} is the same as 3^n times 3^k, so I can write it as:[sum_{k=1}^infty frac{k}{3^k} cdot sum_{n=k+1}^infty frac{1}{3^n}]Yes, that looks better. Now, the inner sum is a geometric series starting from n = k+1. I remember that the sum of a geometric series starting from n = 0 is 1/(1 - r), but here it starts from n = k+1. So, I need to adjust for that.The sum from n = k+1 to infinity of 1/3^n is the same as the sum from n = 0 to infinity of 1/3^{n + k + 1}, right? Because if I let m = n - (k+1), then when n = k+1, m = 0, and so on. So, that sum is:[sum_{m=0}^infty frac{1}{3^{m + k + 1}} = frac{1}{3^{k+1}} sum_{m=0}^infty left(frac{1}{3}right)^m]And the sum of that geometric series is 1/(1 - 1/3) = 3/2. So, putting it all together:[frac{1}{3^{k+1}} cdot frac{3}{2} = frac{1}{2 cdot 3^k}]So, the inner sum simplifies to 1/(2 * 3^k). Now, plugging that back into our expression, we have:[sum_{k=1}^infty frac{k}{3^k} cdot frac{1}{2 cdot 3^k} = frac{1}{2} sum_{k=1}^infty frac{k}{9^k}]Alright, so now I have a single sum left: the sum from k=1 to infinity of k over 9^k, multiplied by 1/2. I need to compute this sum. I remember that there's a formula for the sum of k x^k from k=1 to infinity, which is x/(1 - x)^2, but let me verify that.Let me denote S as the sum:[S = sum_{k=1}^infty frac{k}{9^k}]So, S is equal to 1/9 + 2/9^2 + 3/9^3 + 4/9^4 + ... and so on. To find S, I can use a trick where I multiply both sides by 9:[9S = 1 + frac{2}{9} + frac{3}{9^2} + frac{4}{9^3} + dots]Now, if I subtract S from 9S, let's see what happens:[9S - S = 8S = 1 + left(frac{2}{9} - frac{1}{9}right) + left(frac{3}{9^2} - frac{2}{9^2}right) + left(frac{4}{9^3} - frac{3}{9^3}right) + dots]Simplifying each term inside the parentheses:[8S = 1 + frac{1}{9} + frac{1}{9^2} + frac{1}{9^3} + dots]Ah, now this is a geometric series! The sum of 1 + 1/9 + 1/9^2 + 1/9^3 + ... is equal to 1/(1 - 1/9) = 9/8. So,[8S = frac{9}{8} implies S = frac{9}{64}]Wait, hold on, 8S = 9/8, so S = (9/8)/8 = 9/64? Hmm, let me check that again.Wait, 8S = 9/8, so S = (9/8) / 8 = 9/(8*8) = 9/64. Yeah, that's correct.So, going back, the sum we had earlier was (1/2) * S, which is (1/2) * (9/64) = 9/128.Therefore, the original double sum is equal to 9/128.Wait, let me just recap to make sure I didn't make any mistakes. I switched the order of summation, recognized the inner sum as a geometric series, summed it up, then dealt with the remaining sum using the formula for the sum of k x^k. Everything seems to check out.So, yeah, the final answer should be 9/128.**Final Answer**The sum is equal to boxed{dfrac{9}{128}}.

💡First, I recognize that at each stop, one-third of the students get off the bus, which means two-thirds of the students remain on the bus after each stop.Starting with 60 students, I'll calculate the number of students remaining after each of the four stops.After the first stop, the number of students remaining is 60 multiplied by 2/3, which equals 40 students.After the second stop, the number of students remaining is 40 multiplied by 2/3, resulting in 80/3 students.After the third stop, the number of students remaining is (80/3) multiplied by 2/3, which equals 160/9 students.Finally, after the fourth stop, the number of students remaining is (160/9) multiplied by 2/3, resulting in 320/27 students.Therefore, after the fourth stop, there are 320/27 students remaining on the bus.

💡To find the greatest common divisor (GCD) of all six-digit integers formed by repeating a three-digit number, let's denote the three-digit number as ( x ). The six-digit number can be expressed as ( 1000x + x = 1001x ).Since 1001 is a common factor for all such six-digit numbers, we need to determine if it is the greatest common divisor. The prime factorization of 1001 is ( 7 times 11 times 13 ).To confirm that 1001 is the GCD, consider two examples:- If ( x = 7 ), the six-digit number is ( 1001 times 7 ).- If ( x = 8 ), the six-digit number is ( 1001 times 8 ).The GCD of ( 1001 times 7 ) and ( 1001 times 8 ) is 1001, as 7 and 8 have no common factors other than 1. This reasoning applies to any three-digit number ( x ), confirming that 1001 is indeed the greatest common divisor.

💡First, I recognize that matrix multiplication is linear, which means I can distribute the matrix (mathbf{N}) over the vector combination (3mathbf{a} - 2mathbf{b}).Next, I apply the linearity property:[mathbf{N}(3mathbf{a} - 2mathbf{b}) = 3mathbf{N}mathbf{a} - 2mathbf{N}mathbf{b}]I substitute the given results of (mathbf{N}mathbf{a}) and (mathbf{N}mathbf{b}):[3mathbf{N}mathbf{a} = 3 begin{pmatrix} 2 -3 end{pmatrix} = begin{pmatrix} 6 -9 end{pmatrix}][2mathbf{N}mathbf{b} = 2 begin{pmatrix} 4 5 end{pmatrix} = begin{pmatrix} 8 10 end{pmatrix}]Finally, I subtract the resulting vectors to find the final answer:[begin{pmatrix} 6 -9 end{pmatrix} - begin{pmatrix} 8 10 end{pmatrix} = begin{pmatrix} -2 -19 end{pmatrix}]

💡Okay, so I have this problem where I need to find the derivative ( y'_x ) given the parametric equations:[left{begin{array}{l}x = ln (tan t) y = frac{1}{sin^2 t}end{array}right.]Hmm, I remember that when dealing with parametric equations, the derivative ( frac{dy}{dx} ) can be found using the chain rule. Specifically, ( frac{dy}{dx} = frac{frac{dy}{dt}}{frac{dx}{dt}} ). So, I need to find the derivatives of both ( y ) and ( x ) with respect to ( t ) and then divide them.Let me start with finding ( frac{dy}{dt} ). The function for ( y ) is ( frac{1}{sin^2 t} ), which can also be written as ( csc^2 t ). I think the derivative of ( csc t ) is ( -csc t cot t ), so using the chain rule, the derivative of ( csc^2 t ) should be ( 2 csc t cdot (-csc t cot t) ). Wait, let me double-check that.Yes, if ( y = csc^2 t ), then ( frac{dy}{dt} = 2 csc t cdot (-csc t cot t) = -2 csc^2 t cot t ). Alternatively, in terms of sine and cosine, that would be ( -2 cdot frac{1}{sin^2 t} cdot frac{cos t}{sin t} = -2 frac{cos t}{sin^3 t} ). Okay, so ( frac{dy}{dt} = -frac{2 cos t}{sin^3 t} ).Now, moving on to ( frac{dx}{dt} ). The function for ( x ) is ( ln (tan t) ). I know that the derivative of ( ln u ) is ( frac{u'}{u} ). So, ( frac{dx}{dt} = frac{d}{dt} [ln (tan t)] = frac{1}{tan t} cdot frac{d}{dt} (tan t) ).The derivative of ( tan t ) is ( sec^2 t ), which is ( frac{1}{cos^2 t} ). Therefore, ( frac{dx}{dt} = frac{1}{tan t} cdot frac{1}{cos^2 t} ). But ( tan t = frac{sin t}{cos t} ), so ( frac{1}{tan t} = frac{cos t}{sin t} ). Putting it all together, ( frac{dx}{dt} = frac{cos t}{sin t} cdot frac{1}{cos^2 t} = frac{1}{sin t cos t} ).Alright, so now I have both derivatives:[frac{dy}{dt} = -frac{2 cos t}{sin^3 t}]and[frac{dx}{dt} = frac{1}{sin t cos t}]To find ( frac{dy}{dx} ), I divide ( frac{dy}{dt} ) by ( frac{dx}{dt} ):[frac{dy}{dx} = frac{-frac{2 cos t}{sin^3 t}}{frac{1}{sin t cos t}} = -frac{2 cos t}{sin^3 t} times frac{sin t cos t}{1}]Simplifying this, the ( sin t ) in the denominator cancels with the ( sin t ) in the numerator, leaving ( sin^2 t ) in the denominator. Similarly, the ( cos t ) in the numerator multiplies with the other ( cos t ), giving ( cos^2 t ). So, we have:[frac{dy}{dx} = -2 frac{cos^2 t}{sin^2 t} = -2 cot^2 t]Wait, let me make sure I didn't make a mistake in the simplification. Starting from:[-frac{2 cos t}{sin^3 t} times frac{sin t cos t}{1}]Multiplying the numerators: ( -2 cos t times sin t cos t = -2 sin t cos^2 t ).Multiplying the denominators: ( sin^3 t times 1 = sin^3 t ).So, it becomes:[-frac{2 sin t cos^2 t}{sin^3 t} = -2 frac{cos^2 t}{sin^2 t} = -2 cot^2 t]Yes, that looks correct. So, the derivative ( y'_x ) is ( -2 cot^2 t ).I think that's the final answer. Let me just recap the steps to ensure I didn't skip anything:1. Expressed both ( x ) and ( y ) in terms of ( t ).2. Found ( frac{dy}{dt} ) using the chain rule, resulting in ( -frac{2 cos t}{sin^3 t} ).3. Found ( frac{dx}{dt} ) by differentiating ( ln (tan t) ), which gave ( frac{1}{sin t cos t} ).4. Divided ( frac{dy}{dt} ) by ( frac{dx}{dt} ) to get ( frac{dy}{dx} ).5. Simplified the expression to ( -2 cot^2 t ).Everything seems to check out. I don't think I made any calculation errors, but it's always good to verify each step.Another way to think about it is to express ( y ) in terms of ( x ) directly, but that might be more complicated since ( x = ln (tan t) ) implies ( tan t = e^x ), so ( t = arctan(e^x) ). Then, ( y = frac{1}{sin^2 t} ), which can be written as ( 1 + cot^2 t ), but that might not necessarily make differentiation easier. So, using the parametric derivative approach seems more straightforward.Alternatively, I could express ( y ) in terms of ( x ) using trigonometric identities. Since ( x = ln (tan t) ), then ( e^x = tan t ), so ( sin t = frac{e^x}{sqrt{1 + e^{2x}}} ) and ( cos t = frac{1}{sqrt{1 + e^{2x}}} ). Then, ( y = frac{1}{sin^2 t} = 1 + e^{2x} ). Wait, is that correct?Let me see:If ( tan t = e^x ), then ( sin t = frac{e^x}{sqrt{1 + e^{2x}}} ) and ( cos t = frac{1}{sqrt{1 + e^{2x}}} ). Therefore, ( sin^2 t = frac{e^{2x}}{1 + e^{2x}} ), so ( y = frac{1}{sin^2 t} = frac{1 + e^{2x}}{e^{2x}} = 1 + e^{-2x} ).Wait, that's different from what I initially thought. So, if ( y = 1 + e^{-2x} ), then ( frac{dy}{dx} = -2 e^{-2x} ). But that contradicts the earlier result of ( -2 cot^2 t ). Hmm, so which one is correct?Wait, perhaps I made a mistake in expressing ( y ) in terms of ( x ). Let me double-check.Given ( x = ln (tan t) ), so ( e^x = tan t ). Then, ( sin t = frac{tan t}{sqrt{1 + tan^2 t}} = frac{e^x}{sqrt{1 + e^{2x}}} ). Therefore, ( sin^2 t = frac{e^{2x}}{1 + e^{2x}} ), so ( y = frac{1}{sin^2 t} = frac{1 + e^{2x}}{e^{2x}} = 1 + e^{-2x} ). So, ( y = 1 + e^{-2x} ), which means ( frac{dy}{dx} = -2 e^{-2x} ).But earlier, using the parametric approach, I got ( frac{dy}{dx} = -2 cot^2 t ). So, are these two expressions equivalent?Let me see. Since ( e^{-2x} = frac{1}{e^{2x}} ), and ( e^{x} = tan t ), so ( e^{-2x} = frac{1}{tan^2 t} = cot^2 t ). Therefore, ( -2 e^{-2x} = -2 cot^2 t ). So, both methods give the same result. That’s reassuring.Therefore, the derivative ( y'_x ) is indeed ( -2 cot^2 t ).I think I've covered all the steps and verified the result using two different methods, so I'm confident that the answer is correct.**Final Answer**The derivative ( y'_x ) is boxed{-2 cot^2 t}.

💡First, I need to determine the interest rate per annum using the given information.I know the formula for simple interest is:[ I = P times R times T ]Where:- ( I ) is the interest,- ( P ) is the principal amount,- ( R ) is the rate of interest per annum,- ( T ) is the time in years.I have the following values:- The amount (A) after 2 2/5 years is Rs. 1120,- The principal (P) is Rs. 958.9041095890411,- The time (T) is 2.4 years.First, I'll calculate the interest earned by subtracting the principal from the amount:[ I = A - P = 1120 - 958.9041095890411 = 161.0958904109589 ]Next, I'll use the simple interest formula to solve for the rate (R):[ 161.0958904109589 = 958.9041095890411 times R times 2.4 ]To isolate R, I'll divide both sides by the product of the principal and the time:[ R = frac{161.0958904109589}{958.9041095890411 times 2.4} ]Calculating the denominator:[ 958.9041095890411 times 2.4 = 2301.3698630136985 ]Now, divide the interest by this value:[ R = frac{161.0958904109589}{2301.3698630136985} approx 0.07 ]Finally, to express the rate as a percentage, I'll multiply by 100:[ R = 0.07 times 100 = 7% ]Therefore, the interest rate per annum is 7%.