Question #145876
Let us denote Sn = (a^n) + (b^n) +( c^n) for arbitrary numbers a, b, c. It is known that S1 = 8,5, S2 = 74, 25, S3 = 639, 625 for some values of a, b, c. What is the largest possible value of (S811)^2 - S810.S812?
1
Expert's answer
2020-11-26T16:58:57-0500

S8112S810S812=(a811+b811+c811)2(a810+b810+c810)(a812+b812+c812)=a1622+b1622+c1622+2a811b811+2a811c811++2b811c811a1622b1622c1622a810(b812+c812)b810(a812+c812)c810(a812+b812)==2a811b811+2a811c811+2b811c811a810(b812+c812)b810(a812+c812)c810(a812+b812){a+b+c=8.5a2+b2+c2=74.25a3+b3+c3=639.625since the system is symmetric there will be only one solution:a=0,b=1743054,c=174+30542(1743054)811(174+3054)811(1743054)810(174+3054)812(174+3054)810(1743054)812==(1743054)810(174+3054)810((1743054)2+2(1743054)(174+3054)+(174+3054)2)==(28930516)811(2894)=(1)811(2894)=72.25answer:72.25S_{811}^2 - S_{810}*S_{812} = (a^{811} + b^{811} +c^{811})^2 -\\ (a^{810} + b^{810} +c^{810})*(a^{812} + b^{812} +c^{812}) = \\ a^{1622} + b^{1622} +c^{1622} +2a^{811}b^{811} + 2a^{811}c^{811} +\\ +2b^{811}c^{811} -a^{1622} - b^{1622} -c^{1622} - a^{810}(b^{812} +c^{812})-\\ -b^{810}(a^{812} +c^{812})-c^{810}(a^{812} +b^{812}) = \\ =2a^{811}b^{811} + 2a^{811}c^{811} +2b^{811}c^{811} - a^{810}(b^{812} +c^{812})-\\ -b^{810}(a^{812} +c^{812})-c^{810}(a^{812} +b^{812}) \\ \begin{cases} a+b+c = 8.5\\ a^2+b^2+c^2 = 74.25\\ a^3+b^3+c^3 = 639.625 \end{cases}\\ \text{since the system is symmetric there will be only one solution}:\\ a = 0 , b= \dfrac{17}{4} -\dfrac{\sqrt{305}}{4} , c= \dfrac{17}{4} +\dfrac{\sqrt{305}}{4}\\ 2(\dfrac{17}{4} -\dfrac{\sqrt{305}}{4})^{811}(\dfrac{17}{4} +\dfrac{\sqrt{305}}{4})^{811} - \\ -(\dfrac{17}{4} -\dfrac{\sqrt{305}}{4})^{810}(\dfrac{17}{4} +\dfrac{\sqrt{305}}{4})^{812} - \\ -(\dfrac{17}{4} +\dfrac{\sqrt{305}}{4})^{810}(\dfrac{17}{4} -\dfrac{\sqrt{305}}{4})^{812} = \\ = -(\dfrac{17}{4} -\dfrac{\sqrt{305}}{4})^{810}(\dfrac{17}{4} +\dfrac{\sqrt{305}}{4})^{810}*\\ *((\dfrac{17}{4} -\dfrac{\sqrt{305}}{4})^{2} +2(\dfrac{17}{4} -\dfrac{\sqrt{305}}{4})(\dfrac{17}{4} +\dfrac{\sqrt{305}}{4}) + (\dfrac{17}{4} +\dfrac{\sqrt{305}}{4})^{2}) = \\ = - (\dfrac{289-305}{16})^{811} * (\dfrac{289}{4}) = -(-1)^{811} *(\dfrac{289}{4}) = 72.25\\ answer: 72.25


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS