100If a,-n(n!), then Σ a, is equal to(1) 101!(3) 101! - 110.r-1(2) 10

1.	100If a,-n(n!), then Σ a, is equal to(1) 101!(3) 101! - 110.r-1(2) 100!-1(4) 101! +1
Answer» n(n!) = (n+1-1)n! = (n+1)n! -(1)n! = (n+1)!-n! The terms will become a1 = 2!-1! a2 = 3!-2! a3 = 4!-3!.... and so ona100 = 101!-100! so, on summation it will be ,a1+a2+a3 ..... +a100 = 2!-1!+3!-2!+4!-3!+5!-4!..... +100!-99!+101!-100! here all the negetive terms will cancel its precedding positive terms like 2!-2!, 3!-3! and so on 100!-100! so, at the end the terms remaining will be 101!-1! = 101!-1 so, 101!-1 is the answer. option 3

100If a,-n(n!), then Σ a, is equal to(1) 101!(3) 101! - 110.r-1(2) 100!-1(4) 101! +1

Answer»

n(n!) = (n+1-1)n! = (n+1)n! -(1)n! = (n+1)!-n!

The terms will become

a1 = 2!-1! a2 = 3!-2! a3 = 4!-3!.... and so ona100 = 101!-100!

so, on summation it will be ,a1+a2+a3 ..... +a100 = 2!-1!+3!-2!+4!-3!+5!-4!..... +100!-99!+101!-100!

here all the negetive terms will cancel its precedding positive terms like 2!-2!, 3!-3! and so on 100!-100!

so, at the end the terms remaining will be 101!-1! = 101!-1

so, 101!-1 is the answer.

option 3