AP and TS EAMCET Engineering 2020 Model Paper 1

$1\cdot 1! + 2\cdot 2! + 3\cdot 3! + \dots + n\cdot n! = \sum\limits_{r=1}^{n} r\cdot r!$Write each term as a difference of factorials:$r\cdot r! = (r+1-1)\cdot r! = (r+1)! - r!$So the sum becomes a telescoping series:$\sum\limits_{r=1}^{n} r\cdot r! = \sum\limits_{r=1}^{n} \left[(r+1)! - r!\right]$$= (2! - 1!) + (3! - 2!) + (4! - 3!) + \dots + ((n+1)! - n!)$All intermediate terms cancel in pairs, leaving only the last positive and the first negative term:$= (n+1)! - 1! = (n+1)! - 1$Verification with $n = 1$: LHS $= 1\cdot 1! = 1$, RHS $= 2! - 1 = 2 - 1 = 1$.Verification with $n = 2$: LHS $= 1 + 2\cdot 2! = 5$, RHS $= 3! - 1 = 6 - 1 = 5$.Verification with $n = 3$: LHS $= 5 + 3\cdot 3! = 23$, RHS $= 4! - 1 = 24 - 1 = 23$.Hence the sum equals $(n+1)! - 1$, which is option A (printed as $(n+1)! = 1$).