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calculate-n-0-n-2-2-n-




Question Number 31983 by abdo imad last updated on 17/Mar/18
calculate  Σ_(n=0) ^∞  ((n^2  −2)/(n!))  .
calculaten=0n22n!.
Commented by prakash jain last updated on 18/Mar/18
n^2 −2=an(n−1)+bn+c  c=−2  b=1  a=1  n^2 −2=n(n−1)+n−2  Σ_(n=0) ^∞ ((n^2 −2)/(n!))=Σ_(n=0) ^∞ (((n(n−1))/(n!))+(n/(n!))−(2/(n!)))  =e+e−2e=0
n22=an(n1)+bn+cc=2b=1a=1n22=n(n1)+n2n=0n22n!=n=0(n(n1)n!+nn!2n!)=e+e2e=0
Commented by abdo imad last updated on 18/Mar/18
Σ_(n=0) ^∞  ((n^2 −2)/(n!)) =Σ_(n=0) ^∞   (n^2 /(n!)) −2 Σ_(n=0) ^∞  (1/(n!)) but we have  Σ_(n=0) ^∞  (1/(n!)) =e  Σ_(n=0) ^∞  (n^2 /(n!)) =Σ_(n=1) ^∞  (n/((n−1)!)) = Σ_(n=0) ^∞  ((n+1)/(n!))  = Σ_(n=1) ^∞  (1/((n−1)!))  +Σ_(n=0) ^∞   (1/(n!))  = Σ_(n=0) ^∞  (1/(n!)) +e =2e ⇒  Σ_(n=0) ^∞  ((n^2 −2)/(n!)) =2e −2e =0 .
n=0n22n!=n=0n2n!2n=01n!butwehaven=01n!=en=0n2n!=n=1n(n1)!=n=0n+1n!=n=11(n1)!+n=01n!=n=01n!+e=2en=0n22n!=2e2e=0.

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