Question Number 359 by 123456 last updated on 25/Jan/15
![f_n (x)=1+c_0 (n)x+c_1 (n)x^2 +c_2 (n)x^3 c_n (x)=x(x−1)(x−2)∙∙∙(x−n) n∈N ∣f_4 (1)−f_3 (1)∣=?](https://www.tinkutara.com/question/Q359.png)
$${f}_{{n}} \left({x}\right)=\mathrm{1}+{c}_{\mathrm{0}} \left({n}\right){x}+{c}_{\mathrm{1}} \left({n}\right){x}^{\mathrm{2}} +{c}_{\mathrm{2}} \left({n}\right){x}^{\mathrm{3}} \\ $$$${c}_{{n}} \left({x}\right)={x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\centerdot\centerdot\centerdot\left({x}−{n}\right) \\ $$$${n}\in\mathbb{N} \\ $$$$\mid{f}_{\mathrm{4}} \left(\mathrm{1}\right)−{f}_{\mathrm{3}} \left(\mathrm{1}\right)\mid=? \\ $$
Answered by prakash jain last updated on 24/Dec/14
![c_0 (4)=4=4 c_1 (4)=4(4−1)=12 c_2 (4)=4(4−1)(4−2)=24 c_0 (3)=3 c_1 (3)=6 c_2 (3)=6 f_4 (x)=1+c_0 (4)x+c_1 (4)x^2 +c_2 (4)x^3 f_4 (1)=1+4+12+24=41 f_3 (x)=1+c_0 (3)x+c_1 (3)x^2 +c_2 (3)x^3 f_3 (1)=1+3+6+6=16 ∣f_4 (1)−f_3 (1)∣=25](https://www.tinkutara.com/question/Q362.png)
$${c}_{\mathrm{0}} \left(\mathrm{4}\right)=\mathrm{4}=\mathrm{4} \\ $$$${c}_{\mathrm{1}} \left(\mathrm{4}\right)=\mathrm{4}\left(\mathrm{4}−\mathrm{1}\right)=\mathrm{12} \\ $$$${c}_{\mathrm{2}} \left(\mathrm{4}\right)=\mathrm{4}\left(\mathrm{4}−\mathrm{1}\right)\left(\mathrm{4}−\mathrm{2}\right)=\mathrm{24} \\ $$$${c}_{\mathrm{0}} \left(\mathrm{3}\right)=\mathrm{3} \\ $$$${c}_{\mathrm{1}} \left(\mathrm{3}\right)=\mathrm{6} \\ $$$${c}_{\mathrm{2}} \left(\mathrm{3}\right)=\mathrm{6} \\ $$$${f}_{\mathrm{4}} \left({x}\right)=\mathrm{1}+{c}_{\mathrm{0}} \left(\mathrm{4}\right){x}+{c}_{\mathrm{1}} \left(\mathrm{4}\right){x}^{\mathrm{2}} +{c}_{\mathrm{2}} \left(\mathrm{4}\right){x}^{\mathrm{3}} \\ $$$${f}_{\mathrm{4}} \left(\mathrm{1}\right)=\mathrm{1}+\mathrm{4}+\mathrm{12}+\mathrm{24}=\mathrm{41} \\ $$$${f}_{\mathrm{3}} \left({x}\right)=\mathrm{1}+{c}_{\mathrm{0}} \left(\mathrm{3}\right){x}+{c}_{\mathrm{1}} \left(\mathrm{3}\right){x}^{\mathrm{2}} +{c}_{\mathrm{2}} \left(\mathrm{3}\right){x}^{\mathrm{3}} \\ $$$${f}_{\mathrm{3}} \left(\mathrm{1}\right)=\mathrm{1}+\mathrm{3}+\mathrm{6}+\mathrm{6}=\mathrm{16} \\ $$$$\mid{f}_{\mathrm{4}} \left(\mathrm{1}\right)−{f}_{\mathrm{3}} \left(\mathrm{1}\right)\mid=\mathrm{25} \\ $$