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Question Number 33571 by Joel578 last updated on 19/Apr/18

$$f\left(x\right)=x^{20}+a_1{x}^{19}+a_2{x}^{18}+...+a_{20}$$$$\mathrm{If}f\left(1\right)=f\left(2\right)=f\left(3\right)=...=f\left(20\right)$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{value}\mathrm{of}{a}_1?$$

Answered by MJS last updated on 19/Apr/18

$$\mathrm{let}\mathrm{me}\mathrm{downsize}\mathrm{this}$$$$f\left(x\right)=x^2+a_{1}x+a_2$$$$f\left(1\right)=1+a_1+a_2$$$$f\left(2\right)=4+2a_1+a_2$$$$f\left(2\right)-f\left(1\right)=0$$$$3+a_1=0\Rightarrow a_1=-3$$$$$$$$f\left(x\right)=x^3+a_1{x}^2+a_{2}x+a_3$$$$f\left(1\right)=1+a_1+a_2+a_3$$$$f\left(2\right)=8+4a_1+2a_2+a_3$$$$f\left(3\right)=27+9a_1+3a_2+a_3$$$$f\left(3\right)-f\left(2\right)=0$$$$19+5a_1+a_2=0\left(i\right)$$$$f\left(2\right)-f\left(1\right)=0$$$$7+3a_1+a_2=0\left(ii\right)$$$$\left(i\right)-\left(ii\right)$$$$12+2a_1=0$$$$a_1=-6$$$$$$$$f\left(x\right)=x^4+...\Rightarrow$$$$\Rightarrow a_1=-10$$$$$$$$f\left(x\right)=x^5+...\Rightarrow$$$$\Rightarrow a_1=-15$$$$$$$$f\left(x\right)=x^n+...\Rightarrow$$$$\Rightarrow a_1=-\frac{n}{2}(n+1)$$$$$$$$\Rightarrow f\left(x\right)=x^{20}+...\Rightarrow a_1=-210$$

Commented by Joel578 last updated on 20/Apr/18

$$thankyouverymuch$$

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