Solve (2019^{2}-2018^{2})+(2017^2-2016^2)+(1110+(2^2-1^2)

$$( 2019 ^ { 2 } - 2018 ^ { 2 } ) + ( 2017 ^ { 2 } - 2016 ^ { 2 } ) + ( 1110 + ( 2 ^ { 2 } - 1 ^ { 2 } )$$

Evaluate

$$9183$$

Solution Steps

Calculate 2019 to the power of 2 and get 4076361.

$$4076361-2018^{2}+2017^{2}-2016^{2}+1110+2^{2}-1^{2}$$

Calculate 2018 to the power of 2 and get 4072324.

$$4076361-4072324+2017^{2}-2016^{2}+1110+2^{2}-1^{2}$$

Subtract 4072324 from 4076361 to get 4037.

$$4037+2017^{2}-2016^{2}+1110+2^{2}-1^{2}$$

Calculate 2017 to the power of 2 and get 4068289.

$$4037+4068289-2016^{2}+1110+2^{2}-1^{2}$$

Calculate 2016 to the power of 2 and get 4064256.

$$4037+4068289-4064256+1110+2^{2}-1^{2}$$

Subtract 4064256 from 4068289 to get 4033.

$$4037+4033+1110+2^{2}-1^{2}$$

Add 4037 and 4033 to get 8070.

$$8070+1110+2^{2}-1^{2}$$

Add 8070 and 1110 to get 9180.

$$9180+2^{2}-1^{2}$$

Calculate 2 to the power of 2 and get 4.

$$9180+4-1^{2}$$

Calculate 1 to the power of 2 and get 1.

$$9180+4-1$$

Subtract 1 from 4 to get 3.

$$9180+3$$

Add 9180 and 3 to get 9183.

$$9183$$

Factor

$$3\times 3061$$