💡Okay, so I have this problem here: If ( f(n) = 1^{2} + 2^{2} + 3^{2} + ldots + (2n)^{2} ), then the recursive relation between ( f(k+1) ) and ( f(k) ) is __________. Hmm, I need to figure out how ( f(k+1) ) relates to ( f(k) ). Let me think step by step.First, let's understand what ( f(n) ) represents. It's the sum of the squares of the first ( 2n ) natural numbers. So, for example, if ( n = 1 ), then ( f(1) = 1^2 + 2^2 = 1 + 4 = 5 ). If ( n = 2 ), then ( f(2) = 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30 ). Got it.Now, the question is about finding a recursive relation between ( f(k+1) ) and ( f(k) ). That means I need to express ( f(k+1) ) in terms of ( f(k) ). So, let's write out what ( f(k) ) and ( f(k+1) ) are.( f(k) = 1^2 + 2^2 + 3^2 + ldots + (2k)^2 )Similarly,( f(k+1) = 1^2 + 2^2 + 3^2 + ldots + (2k)^2 + (2k+1)^2 + (2k+2)^2 )Wait, so ( f(k+1) ) is just ( f(k) ) plus the next two terms in the sequence, which are ( (2k+1)^2 ) and ( (2k+2)^2 ). That makes sense because when you go from ( k ) to ( k+1 ), you're adding two more terms to the sum.So, if I subtract ( f(k) ) from ( f(k+1) ), I should get those two additional terms. Let me write that out:( f(k+1) - f(k) = (2k+1)^2 + (2k+2)^2 )Therefore, to express ( f(k+1) ) in terms of ( f(k) ), I can rearrange the equation:( f(k+1) = f(k) + (2k+1)^2 + (2k+2)^2 )Hmm, that seems straightforward. Let me check if this makes sense with an example. Let's take ( k = 1 ). Then,( f(1) = 1^2 + 2^2 = 1 + 4 = 5 )( f(2) = 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30 )According to the recursive formula,( f(2) = f(1) + (2*1+1)^2 + (2*1+2)^2 = 5 + 3^2 + 4^2 = 5 + 9 + 16 = 30 )Yes, that works. Let me try another one. Let ( k = 2 ):( f(2) = 30 )( f(3) = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 = 1 + 4 + 9 + 16 + 25 + 36 = 91 )Using the recursive formula:( f(3) = f(2) + (2*2+1)^2 + (2*2+2)^2 = 30 + 5^2 + 6^2 = 30 + 25 + 36 = 91 )Perfect, that matches as well. So, the recursive relation seems correct.Just to make sure I didn't make a mistake, let me think about the general case. If ( f(n) ) is the sum up to ( 2n ), then ( f(n+1) ) would be the sum up to ( 2(n+1) = 2n + 2 ). So, the difference between ( f(n+1) ) and ( f(n) ) is the sum of the squares of the next two numbers after ( 2n ), which are ( 2n + 1 ) and ( 2n + 2 ). Therefore, ( f(n+1) = f(n) + (2n + 1)^2 + (2n + 2)^2 ).Yes, that aligns with what I derived earlier. So, I think I'm confident that the recursive relation is correct.**Final Answer**The recursive relation is boxed{f(k+1) = f(k) + (2k+1)^{2} + (2k+2)^{2}}.

💡To determine the smallest positive integer ( k ) such that ( 2^5 ), ( 3^3 ), and ( 14^2 ) are all factors of the product ( 936 times k ), we start by factorizing 936.First, factorize 936:[936 = 2^3 times 3^1 times 13^1]Next, consider the prime factorizations of the required factors:[2^5 = 2^5][3^3 = 3^3][14^2 = (2 times 7)^2 = 2^2 times 7^2]To ensure that ( 936 times k ) includes all these factors, we need to determine the additional prime factors required:- For ( 2^5 ), since 936 already has ( 2^3 ), we need ( 2^{5-3} = 2^2 ).- For ( 3^3 ), since 936 has ( 3^1 ), we need ( 3^{3-1} = 3^2 ).- For ( 14^2 ), we need ( 2^2 times 7^2 ). The ( 2^2 ) is already accounted for above, so we only need ( 7^2 ).Thus, the smallest ( k ) must be:[k = 2^2 times 3^2 times 7^2 = 4 times 9 times 49 = 1764]

💡First, I need to determine how many times the line (3x + 4y = 12) intersects the circle (x^2 + y^2 = 4). To do this, I'll start by expressing (y) in terms of (x) from the equation of the line.Solving for (y):[3x + 4y = 12 implies 4y = 12 - 3x implies y = 3 - frac{3x}{4}]Next, I'll substitute this expression for (y) into the equation of the circle:[x^2 + left(3 - frac{3x}{4}right)^2 = 4]Expanding the squared term:[x^2 + left(9 - frac{9x}{2} + frac{9x^2}{16}right) = 4]Combining like terms:[x^2 + 9 - frac{9x}{2} + frac{9x^2}{16} = 4][frac{25x^2}{16} - frac{9x}{2} + 5 = 0]Multiplying through by 16 to eliminate the fraction:[25x^2 - 72x + 80 = 0]To determine the number of real solutions, I'll calculate the discriminant of this quadratic equation:[D = (-72)^2 - 4 times 25 times 80 = 5184 - 8000 = -1816]Since the discriminant is negative ((D = -1816)), there are no real solutions. This means the line does not intersect the circle at any point.

💡First, I need to multiply the numbers inside the square root: (2 times 9 times 2).Calculating the multiplication step by step:(2 times 9 = 18),then (18 times 2 = 36).Now, I have (sqrt{36}).Finally, taking the square root of 36 gives me 6.

💡Okay, so I have this problem here with a parabola and a circle. Let me try to figure it out step by step. First, the parabola is given in parametric form: ( C_1 ) has equations ( x = 8t^2 ) and ( y = 8t ). I remember that parametric equations can often be converted into a standard form by eliminating the parameter. Here, the parameter is ( t ). Looking at the equations, if I solve for ( t ) from the second equation, ( y = 8t ), I can express ( t ) as ( t = frac{y}{8} ). Then, I can substitute this into the first equation: ( x = 8t^2 = 8left(frac{y}{8}right)^2 ). Let me compute that: ( 8 times frac{y^2}{64} = frac{y^2}{8} ). So, the equation becomes ( x = frac{y^2}{8} ), which I can rewrite as ( y^2 = 8x ). Alright, so ( C_1 ) is a parabola opening to the right with vertex at the origin. Now, I need to find its focus. For a parabola in the form ( y^2 = 4ax ), the focus is at ( (a, 0) ). Comparing this with my equation ( y^2 = 8x ), I see that ( 4a = 8 ), so ( a = 2 ). Therefore, the focus of ( C_1 ) is at ( (2, 0) ). Got that down.Next, the circle ( C_2 ) is given in polar coordinates as ( rho = r ), where ( r > 0 ). I know that in polar coordinates, ( rho ) represents the radius, so ( rho = r ) is a circle centered at the origin with radius ( r ). Converting this to Cartesian coordinates, the equation would be ( x^2 + y^2 = r^2 ). That seems straightforward.Now, the problem states that there's a line with slope 1 that passes through the focus of the parabola ( C_1 ) and is tangent to the circle ( C_2 ). I need to find the value of ( r ).Let me break this down. First, the line has a slope of 1 and passes through the focus ( (2, 0) ). The general equation of a line with slope ( m ) passing through a point ( (x_0, y_0) ) is ( y - y_0 = m(x - x_0) ). Plugging in the values, the equation becomes ( y - 0 = 1 times (x - 2) ), which simplifies to ( y = x - 2 ). Alternatively, I can write this in standard form: ( x - y - 2 = 0 ). This will be useful later when calculating distances.Since this line is tangent to the circle ( C_2 ), the distance from the center of the circle to the line must be equal to the radius ( r ). The circle is centered at the origin ( (0, 0) ), so I need to find the distance from ( (0, 0) ) to the line ( x - y - 2 = 0 ).The formula for the distance ( d ) from a point ( (x_0, y_0) ) to the line ( ax + by + c = 0 ) is:[d = frac{|ax_0 + by_0 + c|}{sqrt{a^2 + b^2}}]In this case, the line is ( x - y - 2 = 0 ), so ( a = 1 ), ( b = -1 ), and ( c = -2 ). The point is ( (0, 0) ), so substituting into the formula:[d = frac{|1 times 0 + (-1) times 0 - 2|}{sqrt{1^2 + (-1)^2}} = frac{|-2|}{sqrt{1 + 1}} = frac{2}{sqrt{2}} = sqrt{2}]So, the distance from the center of the circle to the line is ( sqrt{2} ). Since the line is tangent to the circle, this distance must equal the radius ( r ). Therefore, ( r = sqrt{2} ).Let me double-check my steps to make sure I didn't make a mistake. I converted the parametric equations to standard form correctly, found the focus at ( (2, 0) ), wrote the equation of the line with slope 1 passing through that point, converted it to standard form, and then used the distance formula. The calculations seem correct, and the logic follows. So, I think ( r = sqrt{2} ) is the right answer.**Final Answer**boxed{C}

💡First, I need to find the distance from the origin to the given line ( x + sqrt{3}y - 2 = 0 ).I recall that the formula for the distance from a point ( (x_0, y_0) ) to a line ( ax + by + c = 0 ) is:[d = frac{|ax_0 + by_0 + c|}{sqrt{a^2 + b^2}}]In this case, the origin is ( (0, 0) ), and the line is ( x + sqrt{3}y - 2 = 0 ). So, ( a = 1 ), ( b = sqrt{3} ), and ( c = -2 ).Substituting the values into the formula:[d = frac{|1 cdot 0 + sqrt{3} cdot 0 - 2|}{sqrt{1^2 + (sqrt{3})^2}} = frac{|-2|}{sqrt{1 + 3}} = frac{2}{2} = 1]Therefore, the distance from the origin to the line is 1.