Question

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Answer 1

Answer:

option (3)

Step-by-step explanation:

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Answer 2

Given Question :-

Find the value of

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty}\bigg[1 + \dfrac{1}{a_1} \bigg]\bigg[1 + \dfrac{1}{a_2} \bigg] - - - - \bigg[1 + \dfrac{1}{a_n} \bigg]$

$\rm \:a_n = n(a_{n - 1} + 1) \: for \: n \geqslant 2\: and \: a_1 = 1$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:a_1 = 1$

and

$\rm :\longmapsto\:a_n = n(a_{n - 1} + 1)$

For n = 2, we get

$\rm :\longmapsto\:a_2 = 2(a_{1} + 1)$

$\rm \implies\:a_1 + 1 = \dfrac{a_2}{2}$

For n = 3, we get

$\rm :\longmapsto\:a_3 = 3(a_{2} + 1)$

$\rm \implies\:a_2 + 1 = \dfrac{a_3}{3}$

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For n = n + 1, we get

$\rm :\longmapsto\:a_{n + 1} = (n + 1)(a_{n} + 1)$

$\rm \implies\:a_n + 1 = \dfrac{a_{n + 1}}{n + 1}$

Now, Consider

$\rm :\longmapsto\:\bigg[1 + \dfrac{1}{a_1} \bigg]\bigg[1 + \dfrac{1}{a_2} \bigg] - - - - \bigg[1 + \dfrac{1}{a_n} \bigg]$

can be rewritten as

$\rm \:  =  \: \bigg[\dfrac{a_1 + 1}{a_1} \bigg]\bigg[\dfrac{a_2 + 1}{a_2} \bigg] - - - - \bigg[\dfrac{a_n + 1}{a_n} \bigg]$

$\rm \:  =  \: \bigg[\dfrac{a_2}{2} \bigg]\bigg[\dfrac{a_3}{3a_2} \bigg] - - - - \bigg[\dfrac{a_{n + 1}}{(n + 1)a_n} \bigg]$

$\rm \:  =  \: \dfrac{a_{n + 1}}{2.3.4 - - - - (n + 1)}$

$\rm \:  =  \: \dfrac{(n + 1)(a_n + 1)}{2.3.4 - - - - n(n + 1)}$

$\rm \:  =  \: \dfrac{a_n + 1}{2.3.4 - - - - n}$

$\rm \:  =  \: \dfrac{a_n + 1}{n!}$

$\rm \:  =  \: \dfrac{a_n}{n!} + \dfrac{1}{n!}$

$\rm \:  =  \: \dfrac{n(a_{n - 1} + 1)}{n!} + \dfrac{1}{n!}$

$\rm \:  =  \: \dfrac{a_{n - 1} + 1}{(n - 1)!} + \dfrac{1}{n!}$

$\rm \:  =  \: \dfrac{a_{n - 1}}{(n - 1)!} + \dfrac{1}{(n - 1)!} + \dfrac{1}{n!}$

Continuing like this, we get

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$\rm \:  =  \: \dfrac{a_2}{2!} + \dfrac{1}{2!} + \dfrac{1}{3!} + - - + \dfrac{1}{n!}$

$\rm \:  =  \: \dfrac{2(a_1 + 1)}{2!} + \dfrac{1}{2!} + \dfrac{1}{3!} + - - + \dfrac{1}{n!}$

$\rm \:  =  \: \dfrac{2(1+ 1)}{2!} + \dfrac{1}{2!} + \dfrac{1}{3!} + - - + \dfrac{1}{n!}$

$\rm \:  =  \: 2 + \dfrac{1}{2!} + \dfrac{1}{3!} + - - + \dfrac{1}{n!}$

$\rm \:  =  \: 1 + 1 + \dfrac{1}{2!} + \dfrac{1}{3!} + - - + \dfrac{1}{n!}$

$\rm \:  =  \: 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + - - + \dfrac{1}{n!}$

So, Now Consider

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty}\bigg[1 + \dfrac{1}{a_1} \bigg]\bigg[1 + \dfrac{1}{a_2} \bigg] - - - - \bigg[1 + \dfrac{1}{a_n} \bigg]$

$\rm \:  =  \displaystyle\lim_{n \to \infty}\bigg[\: 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + - - + \dfrac{1}{n!} \bigg]$

$\rm \:  =  \: e$

Hence,

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty}\bigg[1 + \dfrac{1}{a_1} \bigg]\bigg[1 + \dfrac{1}{a_2} \bigg] - - - - \bigg[1 + \dfrac{1}{a_n} \bigg] = e$

So, option (a) is correct