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-

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Question Number 359 by 123456 last updated on 25/Jan/15

$$f_n\left(x\right)=1+c_0\left(n\right)x+c_1\left(n\right)x^2+c_2\left(n\right)x^3$$$$c_n\left(x\right)=x(x-1)(x-2)\cdot \cdot \cdot (x-n)$$$$n\in \mathbb{N}$$$$\mid f_4\left(1\right)-f_3\left(1\right)\mid =?$$

Answered by prakash jain last updated on 24/Dec/14

$$c_0\left(4\right)=4=4$$$$c_1\left(4\right)=4(4-1)=12$$$$c_2\left(4\right)=4(4-1)(4-2)=24$$$$c_0\left(3\right)=3$$$$c_1\left(3\right)=6$$$$c_2\left(3\right)=6$$$$f_4\left(x\right)=1+c_0\left(4\right)x+c_1\left(4\right)x^2+c_2\left(4\right)x^3$$$$f_4\left(1\right)=1+4+12+24=41$$$$f_3\left(x\right)=1+c_0\left(3\right)x+c_1\left(3\right)x^2+c_2\left(3\right)x^3$$$$f_3\left(1\right)=1+3+6+6=16$$$$\mid f_4\left(1\right)-f_3\left(1\right)\mid =25$$

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