A) \[\frac{3e}{2}\]
B) \[e\]
C) \[2e\]
D) \[3e\]
Correct Answer: D
Solution :
Let \[S=\frac{2}{1!}+\frac{6}{2!}+\frac{12}{3!}+\frac{20}{4!}+.....\] and let \[{{S}_{1}}=2+6+12+20+.....+{{T}_{n}}\] \[{{S}_{1}}=\,\,\,\,\,\,\,\,\,\,\,2+6+12+.................{{T}_{n-1}}+{{T}_{n}}\] \[\underline{\overline{0=2+4+6+8+......upto\,\,n\,\,terms-{{T}_{n}}}}\] \[{{T}_{n}}=2+4+6+8+.......\text{upto }\,\,n\,\,\text{terms}\] Þ \[{{T}_{n}}\]=\[\frac{n}{2}[2\times 2+(n-1)\,2]\]= \[n(2+n-1)=n(n+1)\] \ nth term of given series \[{{T}_{n}}=\frac{n(n+1)}{n!}\text{or }{{T}_{n}}=\frac{n(n+1)}{n(n-1)!}\] or \[{{T}_{n}}=\frac{1}{(n-2)\,!}+\frac{2}{(n-1)!}\] Now, sum = \[\sum\limits_{n=1}^{\infty }{\frac{1}{(n-2)!}+2\sum\limits_{n=1}^{\infty }{\frac{1}{(n-1)\,!}}}=e+2e=3e\].
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