The value of ( 2 ⋅ 1 P 0 − 3 ⋅ 2 P 1 + 4 ⋅ 3 P 2 − . . . . . . . . up to 51 th term) + ( 1 ! − 2 ! + 3 ! − . . . . . . . up to 51 th term) is equal to

Step-by-step solution

2 · P 0 1 = 2 × 1 ! 1 ! = 2 ! 3 · P 1 2 = 3 × 2 ! 1 ! = 3 ! 4 · P 2 3 = 4 × 3 ! 1 ! = 4 ! By this observation given question can be written as = 2 ! - 3 ! + 4 ! - 5 ! + . . . . . . . - 51 ! + 52 ! + 1 ! - 2 ! + 3 ! - 4 ! + . . . . . . - 50 ! + 51 ! = 1 ! + 52 ! = 1 + 52 !