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Question

The value of $$1 + 1.1! + 2.2! + 3.3! + ..... + n.n!$$ is


A
(n+1)!+1
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B
(n1)!+1
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C
(n+1)!1
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D
(n+1)!
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Solution

The correct option is D $$(n+1)!$$
$$ \displaystyle 1+1.1!+2.2!+3.3!+.....+n.n!\\ \displaystyle =1+\sum _{ r=1 }^{ n }{ r.r! } \\ = \displaystyle 1+\sum _{ r=1 }^{ n }{ \left( r+1-1 \right) r! } \\ \displaystyle =1+\sum _{ r=1 }^{ n }{ \left( r+1 \right) r!-r! } \\ \displaystyle =1+\sum _{ r=1 }^{ n }{ \left( r+1 \right) !-r! } \\ \displaystyle =1+\left( 2!-1! \right) +\left( 3!-2! \right) +\left( 4!-3! \right) +.....+\left( \left( n+1 \right) !-n! \right) \\ \displaystyle =\left( n+1 \right) !$$

Mathematics

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