Question Number 65134 by turbo msup by abdo last updated on 25/Jul/19

Commented by ~ À ® @ 237 ~ last updated on 25/Jul/19
![Let U_n = n!V_n now let look for V_n the equality gives n!V_(n ) −n[(n−1)V_(n−1) ]=−2 then ∀ k >1 V_k −V_(k−1) = ((−2)/(k!)) then Σ_(k=1) ^n (V_k −V_(k−1) )= Σ_(k=1) ^n ((−2)/(k!)) finally V_n = V_0 −Σ_(k=1) ^n (2/(k!)) V_0 =U_0 =a so U_n = a.n! −n!Σ_(k=1) ^n (2/(k!))](https://www.tinkutara.com/question/Q65157.png)
Commented by ~ À ® @ 237 ~ last updated on 25/Jul/19
![Let U_n = n!V_n now let look for V_n the equality gives n!V_(n ) −n[(n−1)V_(n−1) ]=−2 then ∀ k >1 V_k −V_(k−1) = ((−2)/(k!)) then Σ_(k=1) ^n (V_k −V_(k−1) )= Σ_(k=1) ^n ((−2)/(k!)) finally V_n = V_0 −Σ_(k=1) ^n (2/(k!)) V_0 =U_0 =a so U_n = a.n! −n!Σ_(k=1) ^n (2/(k!))](https://www.tinkutara.com/question/Q65158.png)
Commented by mathmax by abdo last updated on 25/Jul/19

Commented by mathmax by abdo last updated on 26/Jul/19
