Evaluate the value of: \(A=\frac{3}{1!+2!+3!}+\frac{4!}{2!+3!+4!}+......+\frac{n}{(n-2)!+(n-1)!+n!}\).

Suppose that \(\frac{k}{(k-2)!+(k-1)!+k!}=\frac{k}{(k-2)![1+(k-1)+k(k-1)]}=\frac{k}{(k-2)!k^2}=\frac{1}{(k-2)!k}=\frac{k-1}{(k-2)!(k-1)k}=\frac{k-1}{k!}=\frac{1}{(k-1)!}-\frac{1}{k!}\). Therefore, \(A=\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+...+\left(\frac{1}{(n-1)!}-\frac{1}{n!}\right)=\frac{1}{2!}-\frac{1}{n!}=\frac{n!-2}{2n!}\).

 Evaluate the value of: \(A=\frac{3}{1!+2!+3!}+\frac{4!}{2!+3!+4!}+......+\frac{n}{(n-2)!+(n-1)!+n!}\).

Solution:

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