answer:We need to prove that ( x(x+1)(2x+1) ) divides ( (x+1)^{2m} - x^{2m} - 2x - 1 ) where ( m ) is a positive integer. To start, let's denote the given function as: [ f(x) = (x+1)^{2m} - x^{2m} - 2x - 1 ] We know that ( x, x+1, 2x+1 ) are pairwise coprime polynomials. Therefore, it is sufficient to show that each of ( x, x+1, 2x+1 ) divides ( f(x) ). Step 1: Show that ( x ) divides ( f(x) ): Evaluate ( f(x) ) at ( x = 0 ): [ f(0) = (0+1)^{2m} - 0^{2m} - 2(0) - 1 = 1^{2m} - 0 - 0 - 1 = 1 - 1 = 0 ] Thus, ( f(0) = 0 ), which implies that ( x mid f(x) ) since a polynomial evaluated to zero at a particular value indicates it has the corresponding root. Step 2: Show that ( x+1 ) divides ( f(x) ): Evaluate ( f(x) ) at ( x = -1 ): [ f(-1) = ((-1)+1)^{2m} - (-1)^{2m} - 2(-1) - 1 ] [ f(-1) = 0^{2m} - 1 + 2 - 1 = 0 - 1 + 2 - 1 = 0 ] Thus, ( f(-1) = 0 ), which implies that ( x+1 mid f(x) ). Step 3: Show that ( 2x+1 ) divides ( f(x) ): Evaluate ( f(x) ) at ( x = -frac{1}{2} ): [ fleft(-frac{1}{2}right) = left(-frac{1}{2} + 1right)^{2m} - left(-frac{1}{2}right)^{2m} - 2left(-frac{1}{2}right) - 1 ] [ fleft(-frac{1}{2}right) = left(frac{1}{2}right)^{2m} - left(frac{1}{2}right)^{2m} + 1 - 1 = 0 ] Thus, ( fleft(-frac{1}{2}right) = 0 ), which implies that ( 2x+1 mid f(x) ). Conclusion: Given that ( x, x+1, 2x+1 ) are pairwise coprime and each divides ( f(x) ), it follows that their product ( x(x+1)(2x+1) ) divides ( f(x) ). Hence, we have: [ x(x+1)(2x+1) mid (x+1)^{2m} - x^{2m} - 2x - 1 ] Therefore, the statement is proven. [blacksquare]

answer:We start similarly by defining the sequence ( (theta_n) ) where ( theta_0 = arccos(frac{7}{17}) ) and for each ( n ), ( theta_n = 2theta_{n-1} ). This provides [ costheta_0 = frac{7}{17}. ] The recursion relation gives us: [ costheta_n = cos(2theta_{n-1}) = 2cos^2(theta_{n-1}) - 1. ] Thus, ( a_n ) and ( costheta_n ) are identical sequences, implying ( a_n = cos(2^n theta_0) ). We analyze: [ sin^2theta_0 = 1 - left(frac{7}{17}right)^2 = frac{240}{289}, ] yielding ( sintheta_0 = frac{sqrt{240}}{17} = frac{4sqrt{15}}{17} ) since ( theta_0 ) is in the first quadrant. Now consider: [ a_0 a_1 ldots a_{n-1} = costheta_0 cos(2theta_0) ldots cos(2^{n-1}theta_0). ] Multiply the entire product by ( sintheta_0 ): [ frac{4sqrt{15}}{17} a_0 a_1 ldots a_{n-1} = frac{4sqrt{15}}{17} sintheta_0 costheta_0 ldots cos(2^{n-1}theta_0) = frac{1}{2^n} sin(2^ntheta_0). ] Thus: [ |a_0 a_1 ldots a_{n-1}| = frac{1}{2^n} cdot frac{17}{4sqrt{15}} |sin(2^ntheta_0)| le frac{1}{2^n} cdot frac{17}{4sqrt{15}}. ] This gives: [ c le frac{17}{4sqrt{15}}. ] Computing ( 100c ), we use ( sqrt{15} approx 3.872 ): [ 100 cdot frac{17}{4cdot3.872} approx 100 cdot 1.098 approx 110. ] boxed{110}

answer:To compute the dot product of the vectors begin{pmatrix} -4 -1 end{pmatrix} and begin{pmatrix} 6 8 end{pmatrix}, we follow the formula for the dot product, which involves multiplying corresponding elements of the vectors and then summing those products. Thus, we have: [ begin{pmatrix} -4 -1 end{pmatrix} cdot begin{pmatrix} 6 8 end{pmatrix} = (-4) cdot 6 + (-1) cdot 8 ] Breaking down the calculation: [ = -24 + (-8) ] [ = -24 - 8 ] [ = -32 ] Therefore, the dot product of the two vectors is boxed{-32}.

answer:(Ⅰ) According to the problem, we know that S_{n}= frac {1}{3}n^{2}+ frac {2}{3}n (where ninmathbb{N}^{*}). When ngeqslant 2, a_{n}=S_{n}-S_{n-1}= frac {1}{3}n^{2}+ frac {2}{3}n-left[ frac {1}{3}(n-1)^{2}+ frac {2}{3}(n-1)right]= frac {2n+1}{3}; When n=1, a_{1}=S_{1}=1 fits the above formula. Thus, the general formula for the sequence {a_{n}} is a_{n}= frac {2n+1}{3} (where ninmathbb{N}^{*}). (Ⅱ) Since b_{n}=a_{n}a_{n+1}cos [(n+1)pi]=(-1)^{n-1}a_{n}a_{n+1}, Therefore, T_{n}=b_{1}+b_{2}+…+b_{n}=a_{1}a_{2}-a_{2}a_{3}+a_{3}a_{4}-a_{4}a_{5}+…+(-1)^{n-1}a_{n}a_{n+1}. From (Ⅰ), we know that the sequence {a_{n}} is an arithmetic sequence with the first term 1 and common difference frac {2}{3}. ① When n=2m (where minmathbb{N}^{*}), T_{n}=T_{2m}=a_{1}a_{2}-a_{2}a_{3}+a_{3}a_{4}-a_{4}a_{5}+…+(-1)^{2m-1}a_{2m}a_{2m+1} =a_{2}(a_{1}-a_{3})+a_{4}(a_{3}-a_{5})+…+a_{2m}(a_{2m-1}-a_{2m+1})=- frac {4}{3}(a_{2}+a_{4}+…+a_{2m})=- frac {4}{3}× frac {a_{2}+a_{2m}}{2}×m =- frac {1}{9}(8m^{2}+12m)=- frac {1}{9}(2n^{2}+6n); ② When n=2m-1 (where minmathbb{N}^{*}), T_{n}=T_{2m-1}=T_{2m}-(-1)^{2m-1}a_{2m}a_{2m+1} =- frac {1}{9}(8m^{2}+12m)+ frac {1}{9}(16m^{2}+16m+3)= frac {1}{9}(8m^{2}+4m+3)= frac {1}{9}(2n^{2}+6n+7). Therefore, T_{n}= begin{cases} - frac {1}{9}(2n^{2}+6n), & text{if } n text{ is even} frac {1}{9}(2n^{2}+6n+7), & text{if } n text{ is odd} end{cases}. To ensure T_{n}geqslant tn^{2} holds for all ninmathbb{N}^{*}, it suffices to have - frac {1}{9}(2n^{2}+6n)geqslant tn^{2} (when n is a positive even number) always hold, which means - frac {1}{9}(2+ frac {6}{n})geqslant t for n being a positive even number, Therefore, tleqslant - frac {5}{9}. Hence, the range of values for the real number t is boxed{(-infty,- frac {5}{9}]}. (Ⅲ) From a_{n}= frac {2n+1}{3}, we know that every term in the sequence {a_{n}} cannot be an even number. ① If there exists a sequence {a_{n_{k}}} (where kinmathbb{N}^{*}) with the first term a_{1} and common ratio q being 2 or 4, then every term in {a_{n_{k}}} except the first one would be even, thus, there does not exist a sequence {a_{n_{k}}} with the first term a_{1} and an even common ratio; ② When q=1, obviously, such a sequence {a_{n_{k}}} does not exist; when q=3, if there exists a sequence {a_{n_{k}}} with the first term a_{1} and common ratio 3, then a_{n_{1}}=1(n_{1}=1), a_{n_{k}}=3^{k-1}= frac {2n_{k}+1}{3}, n_{k}= frac {3^{k}-1}{2}, which means such a sequence {a_{n_{k}}} exists, and n_{k}= frac {3^{k}-1}{2} (where kinmathbb{N}^{*}).