Question

❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​​​​​

Answer 1

Answer:

1n rase to 5 upon 2n

Step-by-step explanation:

first add numerator and denominator then you get

3n rase to 5 upon 6n then,divide it you will get your

1n rase to 5 upon 2n

Answer 2

Given Question

Evaluate the following

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - - + {n}^{n} }$

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - - + {n}^{n} }$

Consider, Numerator

$\rm :\longmapsto\:n + {n}^{2} + {n}^{3} + - - - + {n}^{n}$

Its form an GP series with first term a, common ratio r and number of terms n respectively.

So, Using Sum of n terms of GP we get

$\rm \:  =  \: \dfrac{n( {n}^{n} - 1)}{n - 1}$

can be further rewritten as

$\rm \:  =  \: \dfrac{{n}^{n} - 1}{\dfrac{n - 1}{n} }$

$\rm \:  =  \: \dfrac{{n}^{n} - 1}{1 - \dfrac{1}{n} }$

Thus,

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

Now, Consider Denominator

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {n}^{n}$

$\rm = {1}^{n} + {2}^{n} + {3}^{n} + - - + {(n - 2)}^{n} + {(n - 1)}^{n} + {n}^{n}$

$\rm =  {n}^{n}\bigg[1 + {\bigg[\dfrac{n - 1}{n} \bigg]}^{n} + {\bigg[\dfrac{n - 2}{n} \bigg]}^{n} + - - - + {\bigg[\dfrac{1}{n} \bigg]}^{n}$

$\rm =  {n}^{n}\bigg[1 + {\bigg[1 - \dfrac{ 1}{n} \bigg]}^{n} + {\bigg[1 - \dfrac{2}{n} \bigg]}^{n} + - - - + {\bigg[\dfrac{1}{n} \bigg]}^{n}$

Now, Consider the given expression

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - - + {n}^{n} }$

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

can be rewritten as

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

We know

$\boxed{\tt{ \displaystyle\lim_{x \to \infty} {\bigg[1 + \dfrac{k}{x} \bigg]}^{x} = {e}^{k} \: }}$

So, using this, we get

$\rm \:  =  \: \dfrac{\dfrac{1 - 0}{1 - 0} }{1 + {e}^{ - 1} + {e}^{ - 2} + - - - - \infty }$

$\rm \:  =  \: \dfrac{1}{1 + {e}^{ - 1} + {e}^{ - 2} + - - - - \infty }$

Now, in denominator its an infinite GP series with First term 1 and common ratio 1/e ( <1), so using sum of infinite GP series we get

$\rm \:  =  \: \dfrac{1}{\dfrac{1}{1 - \dfrac{1}{e} } }$

$\rm \:  =  \: \dfrac{1}{\dfrac{1}{ \dfrac{e - 1}{e} } }$

$\rm \:  =  \: \dfrac{1}{\dfrac{e}{e - 1} }$

$\rm \:  =  \: \dfrac{e - 1}{e}$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - - + {n}^{n} } = \frac{e - 1}{e} \: }}$