Explanation
The value of$$\frac{1}{1!} + \frac{1+2}{2!} + \frac{1+2+3}{3!} + \frac{1+2+3+4}{4!}+\cods$$ upto infinite terms is equal to
Given that,
$$a_1=2$$, $$r=\dfrac{a_{n+1}}{a_n}=\dfrac{1}{3}$$
$$ \sum\nolimits_{r=1}^{20}{{{a}_{r}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}} $$
$$ =\dfrac{2\left[ 1-{{\left( \dfrac{1}{3} \right)}^{20}} \right]}{1-\dfrac{1}{3}} $$ $$[n=20]$$
$$ =\dfrac{2\left[ 1-{{\left( \dfrac{1}{3} \right)}^{20}} \right]}{\dfrac{2}{3}} $$
$$ =3\left[ 1-{{\left( \dfrac{1}{3} \right)}^{20}} \right]$$
Hence this is the correct answer.
Series: $$1+\dfrac{3}{{x}^{1}}+\dfrac{9}{{x}^{2}}+\dfrac{27}{{x}^{3}}+……\infty $$
This is a $$G.P$$
With $$Q A=1$$
$$r=\dfrac{3}{x}$$
$${ s }_{ n }=\dfrac { A }{ 1-r } =\dfrac { 1 }{ 1-\dfrac { 3 }{ x } } \quad \left\{ \because r<1 \right\} $$
$${s}_{n}=\dfrac{1\left(x\right)}{x-3}$$...... 1
Also $${s}_{n}=\dfrac{A}{r-1}=\dfrac{1}{3.1}=\dfrac{x}{3-x}$$......2
$$\left\{ \because r\quad >\quad 1 \right\} $$
From $$1$$ and $$2$$
We infer:
$${s}_{n}=\dfrac{1\left(-x\right)}{x-3}$$ is true for
$$\left\{ 3\quad <\quad x \right\}$$
And
$${s}_{n}=\dfrac{x}{3-x}$$ is true then
$$\left\{ x\quad <\quad 3 \right\}$$
Option $$x\left\{ x<-3\& x>3 \right\} $$
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