JEE Main 2020 (Online) 3rd September Morning ShiftQuestion 1
The value of $(2\cdot{}^1P_0-3\cdot{}^2P_1+4\cdot{}^3P_2-\ldots$ up to $51^{\text{th}}$ term$)+(1!-2!+3!-\ldots$ up to $51^{\text{th}}$ term$)$ is equal to :
Permutation and Combination
Solution
The $k$th term of the first series is $(-1)^{k+1}(k+1)\cdot{}^{k}P_{k-1}=(-1)^{k+1}(k+1)!$, so the first series is $2!-3!+\cdots+52!$. Adding $1!-2!+\cdots+51!$ makes everything telescope: total $=1!+52!=1+52!$. Option (A).