Question

[ankamar]

Cho đa thức 

\(f\left(x\right)=x\cdot\left(x+1\right).\left(x+2\right).\left(ãx+b\right)\)

Xác định a;b để \(f\left(x\right)-f\left(x-1\right)=x.\left(x+1\right)\cdot\left(2x+1\right)\)với mọi \(x\)

Answer 1

[Nguyễn Linh Chi]

\(f\left(x-1\right)=\left(x-1\right)\left(x\right)\left(x+1\right)\left(ax-a+b\right)\)

=> \(f\left(x\right)-f\left(x-1\right)=x\left(x+1\right)\left(2x+1\right)\)mọi x

\(\Leftrightarrow x\left(x+1\right)\left(x+2\right)\left(ax+b\right)-\left(x-1\right)x\left(x+1\right)\left(ax-a+b\right)=x\left(x+1\right)\left(2x+1\right)\)mọi x

\(\Leftrightarrow x\left(x+1\right)\left[\left(x+2\right)\left(ax+b\right)-\left(x-1\right)\left(ax-a+b\right)\right]=x\left(x+1\right)\left(2x+1\right)\)mọi x

\(\Leftrightarrow ax^2+2ax+bx+2b-ax^2+ax-bx+ax-a+b=2x+1\)mọi x

\(\Leftrightarrow4ax+3b-a=2x+1\)

Cân bằng hệ số :

\(\hept{\begin{cases}4a=2\\3b-a=1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=\frac{1}{2}\\b=\frac{1}{2}\end{cases}}\)

Answer 2

[Phạm Ngọc Khương]

a) Ta có $$\begin{aligned} f(x)-f(x-1) & =x(x+1)(x+2)(ax+b)-(x-1)x(x+1)(ax+b) \\ & = 4ax^3+3(a+b)x^2+(3b-a)x \end{aligned}$$
Và $x(x+1)(2x+1)=2x^3+3x^2+x$
Vậy $$4ax^3+3(a+b)x^2+(3b-a)x = 2x^3+3x^2+x \iff \begin{cases} 4a=2 \\ 3(a+b)=3 \\ 3b-a=1 \end{cases} \implies a=b= \dfrac{1}{2}$$

b) Ta có
$$\begin{array}{l}1.2.3= f(1)-f(0) \\ 2.3.5=f(2)-f(1) \\ 3.4.7= f(3)-f(2) \\ ... \\ n(n+1)(2n+1)=f(n)-f(n-1) \end{array}$$
$$\implies S=1.2.3+2.3.5+.....+n(n+1)(2n+1)= f(n-1)-f(0)= \boxed{\dfrac{(n-1)n(n+1)^2}{2}}$$