Okay, the "technical" explanation:The drag force acting on an aircraft can be expressed via the equation Fd = .5 * rho * V² * A * Cd, where rho is air density (in kilograms per metre cubed), V is the velocity (in metres per second), A is the cross-sectional area the drag acts on (in square metres), and Cd is the coefficient of drag (varies according to aircraft and has no units). Keeping in mind the equations of steady, unaccelerated flight (Thrust = Drag, Lift = Weight), it follows that for cruising flight, thrust must equal the force of drag. Since T = D, and drag can also be expressed as D = .5 * rho * V² * A * Cd. Since lift must equal weight, and the equation of lift is .5 * rho * V² * S * Cl (rho = air density, V = velocity, S = wing area, and Cl is the lift coefficient), weight must also equal .5 * rho * V² * S * Cl. If you then divide thrust by weight, you get Cl/Cd, which is essentially lift over drag. If you continue the algebra, the thrust required is equal to weight times drag all divided by lift. This, however, only gives you thrust required as a function of velocity.So going further, and extrapolating to include the height: the power required for an aircraft to stay in straight and level flight is equal to the force of drag times a factor of veloctiy, thus P = .5 * rho * V³ * A * Cd. The velocity at sea level in terms of weight can be expressed as sqrt((2 * weight)/(rho0 * S * Cl) or the square root of two times weight divided by the density of air at sea level times the area of the wing time the coefficient of lift. If we're assuming the coefficients of lift and drag remain constant, and the only variables which change are the density and the velocity, power required for straight and level flight at altitude becomes power required at sea level times (in parentheses) the square root of density at sea level divded by density at altitude. If you were to plot this, you would see that at sea level, for a given altitude, the power required is generally higher than that at a given altitude. Because there is less power required because of decreased drag etc. the thrust to achieve the required power is less. Therefore, less fuel is needed to maintain the forces of flight.The "simple" explanation:Drag is equal to thrust in cruise flight, and since air density decreases as you get higher, and since drag is proportional to both velocity and air density, it means as the air density becomes lower, but the velocity becomes"slightly" higher as you increase altitude, the overal drag force decreases as you climb. Since you're not constantly accelerating as you climb higher (per se, at least), the assumption can be made that the velocity INCREASE is proportionally less than the air density DECREASE, which in turn leads to the above statement. Since the thrust is equal to the drag, and the net force must equal 0 (i.e., the drag equals the thrust, since the forces are applied in opposite directions), the thrust also decreases as the drag decreases. This means that your velocity may slightly increase, but that your drag still decreases, that less thrust is required and thus that your fuel consumption is lowered for the velocity at which you are flying.Quick case and point for the proportionality of velocity increase and air density decrease: you are flying at around 320ktas when you transition from <250KIAS to >250KIAS (i.e. the airspeed transition at 10,000 feet). Your TAS at cruise is generally around 480. The proportion here is 1.5. The air density at 10,000 feet is about .90 kg/m³, and at FL350 about .36 kg/m³ or a proportion of 2.5. Since D = T = constant * rho * V² * constant * constant and rho is decreasing with a linear factor -2.5 (or increasing at a linear factor 0.4), V² increase at a linear factor 2.25, which results in a net decrease (i.e., factor less than 1 -- 0.9 to be exact).To the other guys, jet engines aren't more efficient at higher altitudes, they just need to do less work to provide the same force to overcome drag.References: Introduction to Flight by John Anderson, and Elements of airplane performance by Ger Ruijghok.Also, this is the quick crash-course version. I could get a lot more technical, but I'll spare you. If you're interested, you can always PM me.

Lol.

To the other guys, jet engines aren't more efficient at higher altitudes, they just need to do less work to provide the same force to overcome drag.

There we go.

Okay, the "technical" explanation:The drag force acting on an aircraft can be expressed via the equation Fd = .5 * rho * V² * A * Cd, where rho is air density (in kilograms per metre cubed), V is the velocity (in metres per second), A is the cross-sectional area the drag acts on (in square metres), and Cd is the coefficient of drag (varies according to aircraft and has no units). Keeping in mind the equations of steady, unaccelerated flight (Thrust = Drag, Lift = Weight), it follows that for cruising flight, thrust must equal the force of drag. Since T = D, and drag can also be expressed as D = .5 * rho * V² * A * Cd. Since lift must equal weight, and the equation of lift is .5 * rho * V² * S * Cl (rho = air density, V = velocity, S = wing area, and Cl is the lift coefficient), weight must also equal .5 * rho * V² * S * Cl. If you then divide thrust by weight, you get Cl/Cd, which is essentially lift over drag. If you continue the algebra, the thrust required is equal to weight times drag all divided by lift. This, however, only gives you thrust required as a function of velocity.So going further, and extrapolating to include the height: the power required for an aircraft to stay in straight and level flight is equal to the force of drag times a factor of veloctiy, thus P = .5 * rho * V³ * A * Cd. The velocity at sea level in terms of weight can be expressed as sqrt((2 * weight)/(rho0 * S * Cl) or the square root of two times weight divided by the density of air at sea level times the area of the wing time the coefficient of lift. If we're assuming the coefficients of lift and drag remain constant, and the only variables which change are the density and the velocity, power required for straight and level flight at altitude becomes power required at sea level times (in parentheses) the square root of density at sea level divded by density at altitude. If you were to plot this, you would see that at sea level, for a given altitude, the power required is generally higher than that at a given altitude. Because there is less power required because of decreased drag etc. the thrust to achieve the required power is less. Therefore, less fuel is needed to maintain the forces of flight.The "simple" explanation:Drag is equal to thrust in cruise flight, and since air density decreases as you get higher, and since drag is proportional to both velocity and air density, it means as the air density becomes lower, but the velocity becomes"slightly" higher as you increase altitude, the overal drag force decreases as you climb. Since you're not constantly accelerating as you climb higher (per se, at least), the assumption can be made that the velocity INCREASE is proportionally less than the air density DECREASE, which in turn leads to the above statement. Since the thrust is equal to the drag, and the net force must equal 0 (i.e., the drag equals the thrust, since the forces are applied in opposite directions), the thrust also decreases as the drag decreases. This means that your velocity may slightly increase, but that your drag still decreases, that less thrust is required and thus that your fuel consumption is lowered for the velocity at which you are flying.Quick case and point for the proportionality of velocity increase and air density decrease: you are flying at around 320ktas when you transition from <250KIAS to >250KIAS (i.e. the airspeed transition at 10,000 feet). Your TAS at cruise is generally around 480. The proportion here is 1.5. The air density at 10,000 feet is about .90 kg/m³, and at FL350 about .36 kg/m³ or a proportion of 2.5. Since D = T = constant * rho * V² * constant * constant and rho is decreasing with a linear factor -2.5 (or increasing at a linear factor 0.4), V² increase at a linear factor 2.25, which results in a net decrease (i.e., factor less than 1 -- 0.9 to be exact).To the other guys, jet engines aren't more efficient at higher altitudes, they just need to do less work to provide the same force to overcome drag.References: Introduction to Flight by John Anderson, and Elements of airplane performance by Ger Ruijghok.Also, this is the quick crash-course version. I could get a lot more technical, but I'll spare you. If you're interested, you can always PM me.[EDIT] That last line came out wrong... I don't mean to come over as arrogant, but that's exactly how I sound now that I reread it. Sorry about that. ;)

A little knowledge is a dangerous thing as they say. You've taken no account of induced drag so your formula in this context is fairly useless.