If sum_{r=1}^{10} r !left( r ^{3}+6 r ^{2}+2 | vedclass

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Question

$\sum_{ r =1}^{10} r !\{( r +1)( r +2)( r +3)-9( r +1)+8\}$

$=\sum_{ r =1}^{10}[\{( r +3) !-( r +1) !\}-8\{( r +1) !- r !\}]$

$=(13 !+12 !-2 !-3

Vedclass · Accepted Answer

$160$

Vedclass · Answer

$\sum_{ r =1}^{10} r !\{( r +1)( r +2)( r +3)-9( r +1)+8\}$

$=\sum_{ r =1}^{10}[\{( r +3) !-( r +1) !\}-8\{( r +1) !- r !\}]$

$=(13 !+12 !-2 !-3 !)-8(11 !-1)$

$=(12.13+12-8) \cdot 11 !-8+8$

$=(160)(11) !$

Hence $\alpha=160$

If $\sum_{r=1}^{10} r !\left( r ^{3}+6 r ^{2}+2 r +5\right)=\alpha(11 !),$ then the value of $\alpha$ is equal to ...... .

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