Cho đa thức $A(x)=\left(2x+1\right)^{50}=a_{50}x^{50}+a_{49}x^{49}+a_{48}x^{48}\ +\,.\!.\!.+\ a_1x+ a_0$. Tính $S=a_{50}+a_{49}+a_{48}\ +\,.\!.\!.+\ a_1+a_0$.

Answer

`A(x) = (2x + 1)^{50} = a_{50}x^{50} + a_{49}x^{49} + a_{48}x^{48} + ... + a_{1}x + a_{0}`

Xét `A(x) = A(0)`

`=> A(0) = (2 . 0 + 1)^{50} = a_{50} . 0^{50} + a_{49} . 0^{49} + a_{48} . 0^{48} + ... + a_{1} . 0 + a_{0}`

`=> A(0) = 1^{50} = 0`

`=> A(0) = 1 = 0` (KTM)

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Xét `A(x) = A(1)` 

`=> A(1) = (2 . 1 + 1)^[50] = a_{50} . 1^{50} + a_{49} . 1^{49} + a_{48} . 1^{48} + ... + a_{1} . 1 + a_{0}`

`=> A(1) = (2 + 1)^[50] = a_{50} . 1 + a_{49} . 1 + a_{48} . 1 + ... + a_{1} . 1 + a_{0}`

`=> A(1) = 3^{50} = a_{50} + a_{49} + a_{48} + ... + a_{1} + a_{0}`

`=> A(1) = 3^{50} = S` 

Vậy `S = 3^{50}`