Cho \(f\left( x \right) = {x^{99}} - 101{x^{98}} + 101{x^{97}} - 101{x^{96}} + ... + 101x - 1\). Tính \(f\left( {100} \right).\)

Ta có: \(f\left( x \right) = {x^{99}} - 101{x^{98}} + 101{x^{97}} - 101{x^{96}} + ... + 101x - 1\)

\( = {x^{99}} - (100 + 1){x^{98}} + (100 + 1){x^{97}} - ... - (100 + 1){x^2} + (100 + 1)x - 1\)

\( = {x^{99}} - 100{x^{98}} - {x^{98}} + 100{x^{97}} + ... -100x^2 - {x^2} + 100x + x - 1\)

\( = ({x^{99}} - 100{x^{98}}) - ({x^{98}} - 100{x^{97}}) + ... - ({x^2} - 100x) + x - 1\)

Thay \(x = 100\) vào \(f(x)\) ta được:

\(f(100) = ({100^{99}} - {100.100^{98}}) - ({100^{98}} - {100.100^{97}}) + ... - ({100^2} - 100.100) + 100 - 1\)

\( = ({100^{99}} - {100^{99}}) - ({100^{98}} - {100^{98}}) + ... - ({100^2} - {100^2}) + 99 = 99\)

Vậy \(f(100) = 99\).