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Question Number 32987 by NECx last updated on 08/Apr/18

$$If1=1$$$$2=4$$$$3=10$$$$4=20$$$$5=?$$

Answered by MJS last updated on 09/Apr/18

$$\mathrm{one}\mathrm{possibility}:$$$$a_n=\frac{1}{6}(n+2)(n+1)n$$$$a_5=35$$$$$$$$\mathrm{with}k\mathrm{given}\mathrm{points}\mathrm{a}\mathrm{polynome}$$$$\mathrm{function}\mathrm{of}\mathrm{degree}k-1\mathrm{is}\mathrm{determined}$$$$\mathrm{in}\mathrm{this}\mathrm{case}:$$$$y={ax}^3+{bx}^2+cx+d$$$$$$$$1=a+b+c+d$$$$4=8a+4b+2c+d$$$$10=27a+9b+3c+d$$$$20=64a+16b+4c+d$$$$$$$$\mathrm{solving}\mathrm{leads}\mathrm{to}$$$$a=\frac{1}{6};b=\frac{1}{2};c=\frac{1}{3};d=0$$

Commented by MJS last updated on 09/Apr/18

$$\Rightarrow \mathrm{we}\mathrm{can}\mathrm{freely}\mathrm{set}{a}_5\mathrm{and}\mathrm{find}$$$$\mathrm{a}\mathrm{polynome}\mathrm{of}\mathrm{degree}4$$

Commented by MJS last updated on 09/Apr/18

$$\mathrm{if}\mathrm{we}\mathrm{set}$$$$y=e^{{ax}^3+{bx}^2+cx+d}$$$$\mathrm{and}\mathrm{solve}\mathrm{as}\mathrm{above}\mathrm{we}\mathrm{get}$$$$a_5=\frac{1024}{25}$$$$y=\frac{a}{x^3}+\frac{b}{x^2}+\frac{c}{x}+d\Rightarrow a_5=\frac{3627}{125}$$$$...$$

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