A) \[e-2\]
B) \[\frac{2}{3}e-1\]
C) 1
D) 3/2
Correct Answer: C
Solution :
\[{{T}_{n}}=\frac{n}{(n+1)!}=\frac{n+1-1}{(n+1)!}\] = \[\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}=\frac{1}{(n)!}-\frac{1}{(n+1)!}\] \[\therefore {{T}_{1}}+{{T}_{2}}+{{T}_{3}}+.....\] = \[\left( \frac{1}{1!}-\frac{1}{2!} \right)+\left( \frac{1}{2!}-\frac{1}{3!} \right)+\left( \frac{1}{3!}-\frac{1}{4!} \right)+......\] = \[\left( \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...... \right)\]\[-\left( \frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...... \right)\] = \[\left( 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+....-1 \right)\] \[-\left( 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+....-1-1 \right)\] = \[(e-1)-(e-2)\] = 1.
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