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[1.1!+2.2!+3.3!+........+n.n!=],[" 1) "(...

[1.1!+2.2!+3.3!+........+n.n!=],[" 1) "(n+1)!1," 2) "(n-1)!+1]quad " 3) "(n+1)!+1quad " 4) "(n-1)!-1

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The value of 1+1.1!+2.2!+3.3!+.........+n.n! is (A) (n+1)!+1(B)(n-1)!+i(C)(n+1)!-1(D)(n+1)!

1.2+2.3+3.4+.........+n(n+1)=(1)/(3)n(n+1)(n+3)

1.3+3.5+5.7+......+(2n-1)(2n+1)=(n(4n^(2)+6n-1))/(3)

1.2.3+2.3.4+....+n(n+1)(n+2)=(n(n+1)(n+2)(n+3))/4

Prove that ((2n+1)!)/(n !)=2^n{1. 3. 5 .........(2n-1)(2n+1)}

If A=[[1 , 1, 1],[ 1, 1, 1],[ 1, 1 , 1]] , prove that A^n=[[3^(n-1), 3^(n-1) , 3^(n-1)],[ 3^(n-1), 3^(n-1) , 3^(n-1)],[ 3^(n-1) , 3^(n-1), 3^(n-1)]] n in N .

lim_ (n rarr oo) (1) / (n) [(1) / (n + 1) + (2) / (n + 2) + ... + (3n) / (4n)]

If 1*1!+2*2!+3*3!+ . . .+n*n ! =(n+1)!-1 then show that, 1*1!+2*2!+3*3!+ . . . +n*n! +(n+1)(n+1)! =(n+2)!-1