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1.1! + 2.2! + 3.3! + ......... + n.n! = ?

$\begin{array}{1 1}(A) (n+1)! \\ (B) (n+1)!-1 \\ (C) (n+1)!+1 \\ (D) (n+1)!-n! \end{array}$

1 Answer

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  • $(n+1).n!=(n+1)!$
$1.1!+2.2!+3.3!+................n.n!=$
$(2.1!-1!)+(3.2!-2!)+(4.3!-3!)+.........((n+1).n!-n!)$
$=(2.1!+3.2!+4.3!+.......(n+1).n!)\:-\:(1!+2!+3!+.........n!)$
$=(2!+3!+4!+.......(n+1)!)\:-\:(1!+2!+3!+........n!)$
$=(n+1)!-1$
answered Aug 13, 2013 by rvidyagovindarajan_1
 

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