The value of (2.¹P₀ – 3.²P₁ + 4.³P₂ .... up to 51ᵗʰ term)
+ (1! – 2! + 3! – ..... up to 51ᵗʰ term) is equal to :

Solution

Maths sir · 4 steps

Turn each permutation term into a plain factorial, and the two series cancel almost term by term.

1. General term. The rth term of the first bracket is (-1)ʳ⁺¹(r+1)· ʳPᵣ₋₁. Since ʳPᵣ₋₁ = r!/1! = r!, the term becomes (-1)ʳ⁺¹(r+1)· r! = (-1)ʳ⁺¹(r+1)!.
2. Write both out. With r from 1 to 51, the first bracket is 2! - 3! + 4! - … + 52!. The last term is +52! because r = 51 is odd. The second bracket is 1! - 2! + 3! - … + 51!.

   (2! - 3! + … - 51! + 52!) + (1! - 2! + 3! - … + 51!)

3. Cancel pairs. Every factorial from 2! to 51! appears once in each bracket with opposite signs, so they all cancel. Only 1! from the second bracket and 52! from the first are left.
4. So. The sum is 1! + 52! = 1 + (52)!, which is option (d).

The idea to keep. Rewrite every term in one common form, here a single factorial, before adding. Two alternating series that are shifted by one then cancel, and only the two end terms are left.

Answer from the NTA final answer key. Solution by PrepWiser.