Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 199547 by hardmath last updated on 05/Nov/23

Answered by mr W last updated on 05/Nov/23

$$1)$$$$n\times n!=(n+1-1)\times n!=(n+1)!-n!$$$$1\times 1!+2\times 2!+...+2022\times 2022!$$$$=2!-1!+3!-2!+....+2023!-2022!$$$$=2023!-1!$$$$\Rightarrow ab=2023\times 1=2023$$$$$$$$2)$$$$f(a+b)=f\left(a\right)+f\left(b\right)+ab$$$$f\left(2a\right)=2f\left(a\right)+a^2$$$$$$$$f\left(2\right)=2f\left(1\right)+1^2=5\Rightarrow f\left(1\right)=2$$$$f\left(4\right)=2f\left(2\right)+2^2=2\times 5+4=14$$$$f\left(5\right)=f\left(4\right)+f\left(1\right)+4\times 1=14+2+4=20$$$$f\left(10\right)=2f\left(5\right)+5^2=2\times 20+25=65$$$$f\left(11\right)=f\left(10\right)+f\left(1\right)+10\times 1=65+2+10=77$$$$$$$$orgenerallyf\left(x\right)=\frac{x(x+3)}{2}$$$$\Rightarrow f\left(11\right)=\frac{11\times 14}{2}=77$$

Commented by hardmath last updated on 05/Nov/23

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor}\mathrm{cool}$$

Commented by hardmath last updated on 05/Nov/23

$$thankyoudearprofessorperfectsolutions$$

Commented by hardmath last updated on 05/Nov/23

$$\mathrm{dear}\mathrm{professor},$$$$\mathrm{how}\mathrm{did}\mathrm{you}\mathrm{get}\mathrm{the}\mathrm{generalized}\mathrm{solution}$$$$\mathrm{and}\mathrm{is}\mathrm{it}\mathrm{always}\mathrm{true}?$$

Commented by mr W last updated on 05/Nov/23

$$sincef(x+y)=f\left(x\right)+f\left(y\right)+xy,youcan$$$$assumef\left(x\right)={ax}^2+bx+cwithf\left(2\right)=5.$$$$thenyougeta=\frac{1}{2},b=\frac{3}{2},c=0.$$

Answered by deleteduser1 last updated on 05/Nov/23

$$f\left(4\right)=10+4=14\Rightarrow f\left(8\right)=44$$$$5=2f\left(1\right)+1\Rightarrow f\left(1\right)=2;f\left(3\right)=7+2=9$$$$f\left(11\right)=44+9+24=77$$

Commented by hardmath last updated on 05/Nov/23

$$thankyouprofessor$$

Answered by ajfour last updated on 05/Nov/23

$$2.f(x+h)-f\left(x\right)=f\left(h\right)+hx$$$$f{}^{\prime }\left(x\right)=\frac{f\left(0\right)}{h}+x$$$$\Rightarrow f\left(0\right)=0$$$$f\left(x\right)=\frac{x^2}{2}+kx$$$$asf\left(2\right)=\frac{2^2}{2}+2k=5\Rightarrow k=\frac{3}{2}$$$$f\left(x\right)=\frac{x^2}{2}+\frac{3x}{2}=\frac{x(x+3)}{2}$$

Commented by hardmath last updated on 05/Nov/23

$$thankyouprofessor$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com