answer:To prove the theorem, we need to show that for any integer ( n ), the following equality holds: leftlfloor frac{n}{2} rightrfloor + leftlfloor frac{n+1}{2} rightrfloor = n. We will handle two cases: ( n ) being even and ( n ) being odd. 1. **Case 1: ( n ) is even** Let ( n = 2m ) for some integer ( m ). - First, we evaluate ( leftlfloor frac{n}{2} rightrfloor ): leftlfloor frac{n}{2} rightrfloor = leftlfloor frac{2m}{2} rightrfloor = leftlfloor m rightrfloor = m - Next, we evaluate ( leftlfloor frac{n+1}{2} rightrfloor ): leftlfloor frac{n+1}{2} rightrfloor = leftlfloor frac{2m+1}{2} rightrfloor Since ( 2m+1 ) is an odd number, ( frac{2m+1}{2} ) is ( m + 0.5 ), and the floor function of this is: leftlfloor m + 0.5 rightrfloor = m - Summing these results gives: leftlfloor frac{n}{2} rightrfloor + leftlfloor frac{n+1}{2} rightrfloor = m + m = 2m = n 2. **Case 2: ( n ) is odd** Let ( n = 2m + 1 ) for some integer ( m ). - First, we evaluate ( leftlfloor frac{n}{2} rightrfloor ): leftlfloor frac{n}{2} rightrfloor = leftlfloor frac{2m+1}{2} rightrfloor Since ( 2m+1 ) is an odd number, ( frac{2m+1}{2} ) is ( m + 0.5 ), and the floor function of this is: leftlfloor m + 0.5 rightrfloor = m - Next, we evaluate ( leftlfloor frac{n+1}{2} rightrfloor ): leftlfloor frac{n+1}{2} rightrfloor = leftlfloor frac{2m+2}{2} rightrfloor = leftlfloor m + 1 rightrfloor = m + 1 - Summing these results gives: leftlfloor frac{n}{2} rightrfloor + leftlfloor frac{n+1}{2} rightrfloor = m + (m + 1) = 2m + 1 = n In conclusion, we have shown that for any integer ( n ), whether even or odd, the given equality holds. Therefore, boxed{leftlfloor frac{n}{2} rightrfloor + leftlfloor frac{n+1}{2} rightrfloor = n}

answer:First, recognize 2401x^4+16 as a sum of fourth powers. We can write 2401x^4+16 as (7x)^4 + 2^4. The formula for the sum of fourth powers is: a^4 + b^4 = (a^2 + b^2)(a^2 - b^2 + 2ab)(a^2 - b^2 - 2ab) However, a simpler form without multiple binomials that still uses the identity for educational purposes: a^4 + b^4 = (a+b)(a^3-a^2b+ab^2-b^3) Thus, applying the identity: (7x)^4 + 2^4 = (7x + 2)((7x)^3 - (7x)^2 cdot 2 + 7x cdot 2^2 - 2^3) = (7x + 2)(343x^3 - 98x^2 + 28x - 8) Therefore, a+b+c+d+e+f = 7 + 2 + 343 - 98 + 28 - 8 = boxed{274}.

answer:(1) From the given conditions, we have the following system of equations: begin{cases} f(0)=b+c=0 f( frac {pi}{3})= frac {sqrt {3}}{2}a+frac {b}{2}+c=1 f'( frac {pi}{3})= frac {a}{2}- frac {sqrt {3}}{2}b=0end{cases} Solving the system of equations yields a=sqrt{3}, b=1, and c=-1. Therefore, the function can be expressed as: f(x)=sqrt{3}sin x+cos x-1. (2) From part (1), we know that a=sqrt{3}b and c=-b. Therefore, frac{b}{a}=frac{sqrt{3}}{3}quadtext{and}quadfrac{c}{a}=-frac{sqrt{3}}{3}. Now, let's compare f(frac{b}{a}) and f(frac{c}{a}): fleft(frac{b}{a}right)-fleft(frac{c}{a}right)=2asinleft(frac{sqrt{3}}{3}right). Since a>0 and sinleft(frac{sqrt{3}}{3}right)>0, we have: fleft(frac{b}{a}right)-fleft(frac{c}{a}right)>0, which implies that: boxed{fleft(frac{b}{a}right)>fleft(frac{c}{a}right)}.

answer:1. **Identify Circumcenters**: - P is the circumcenter of triangle ABE. - Q is the circumcenter of triangle BCE. - R is the circumcenter of triangle CDE. - S is the circumcenter of triangle ADE. 2. **Properties of a Kite**: - The diagonals of a kite are perpendicular. Therefore, AC is perpendicular to BD at E. - Since AB = AD and BC = DC, the triangles ABE and ADE, and BCE and CDE are congruent respectively. 3. **Congruence of Triangles**: - Due to the congruence of the triangles, the circumcenters P and S, and Q and R are symmetrically placed about line AC and line BD respectively. 4. **Relationships in PQRS**: - Line PS is perpendicular to AC and line QR is perpendicular to BD. - Due to symmetry and perpendicularity, PS and QR are parallel and equal in length. 5. **Conclude the Nature of PQRS**: - From the properties of a kite and the symmetry of circumcenters, PQRS is a rectangle (as opposite sides are equal and parallel). Therefore, the conclusion is that PQRStext{ is a rectangle}. The final answer is boxed{text{B}}.