Question

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Kinda Easy Question

$\huge \displaystyle\sf\lim\limits_{x\to\infty}\sqrt[x]{\dfrac{x!}{x^{x}}}$Or$\displaystyle\sf\lim\limits_{x\to\infty}\left(\dfrac{x!}{x}\right)^{\left(\dfrac{1}{x}\right)}$​​​

Answer 1

$\huge \boxed{ \tt✠Answer}$

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$\footnotesize \sf\longrightarrow {Assume\:\:\displaystyle \lim_{x\to\infty}\left ( \dfrac{x!}{x} \right )^{\dfrac{1}{x}} = L}$

$\footnotesize \sf\longrightarrow{ln(L)=\displaystyle \lim_{x\to\infty} \left ( \dfrac{1}{x} \right )ln \left ( \dfrac{x!}{x} \right )}$

Put value of x!

$\footnotesize \sf\longrightarrow{ln(L)=\displaystyle \lim_{x\to\infty} \left ( \dfrac{1}{x} \right )ln \left ( \dfrac{x(x-1)!}{x} \right )}$

$\footnotesize \sf\longrightarrow{ln(L)=\displaystyle \lim_{x\to\infty} \left ( \dfrac{1}{x} \right )ln \left ( (x-1)! \right )}$

$\footnotesize \sf\longrightarrow{ln(L)=\displaystyle \lim_{x\to\infty} \dfrac{ln \left ( (x-1)! \right )}{x}}$

$\pmb{\tt{Multiplying \ with\ ( x - 1 )\ in\ numerator\ and\ denominator\, we\ get\ }}$

$\footnotesize \sf\longrightarrow{ln(L)=\displaystyle \lim_{x\to\infty} (x-1)\dfrac{ln \left ( (x-1)! \right )}{x(x-1)}}$

$\footnotesize \sf\longrightarrow {\displaystyle \lim_{x\to\infty} \dfrac{ln \left ( x! \right )}{x}=\infty}$

$\footnotesize \sf\longrightarrow {ln(L)=(\infty) \displaystyle \lim_{x\to\infty} \dfrac{x-1}{x}}$

$\footnotesize \sf\longrightarrow {ln(L)=(\infty) \displaystyle \lim_{x\to\infty} \left(1-\dfrac{1}{x}\right)}$

$\boxed{\tt{Put \: value \: of \: limits,}}$

$\footnotesize \sf\longrightarrow{ln(L)=(\infty) \left(1-\dfrac{1}{\infty}\right)}$

$\footnotesize \sf\longrightarrow {ln(L)=(\infty) \left(1-0\right)}$

$\footnotesize \sf\longrightarrow {ln(L)=\infty}$

$\footnotesize \sf\longrightarrow{L=e^{\infty}}$

$\huge\boxed{ \tt{L=\infty}}$

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Answer 2

Answer:

Solution:-

The sequence has a common difference of 8, and the first term 4.

The formula for series of an arithmetic progression is,

\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}=\dfrac{n\{2a+(n-1)d\}}{2}↪k=1∑nak=2n{2a+(n−1)d}

where aa is the first term, dd is the common difference, and nn is the term number.

\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}=\dfrac{n\{8+8(n-1)\}}{2}↪k=1∑nak=2n{8+8(n−1)}

\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}=4n^{2}↪k=1∑nak=4n2

\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}=(2n)^{2}↪k=1∑nak=(2n)2

So, the sum of the sequence is always a perfect square.

\large\text{\underline{Maths activity:-}}Maths activity:-

Also, we can make another perfect square sum.

Since,

\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}-\sum^{n-1}_{k=1}a_{k}=a_{n}\ (n\geq2)\text{ and }\sum^{1}_{k=1}a_{k}=a_{1}↪k=1∑nak−k=1∑n−1ak=an (n≥2) and k=1∑1ak=a1

Put a perfect square in terms of nn . We get,

\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}=(kn)^{2}↪k=1∑nak=(kn)2

Then,

\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}-\sum^{n-1}_{k=1}a_{k}=(kn)^{2}-\{k(n-1)\}^{2}↪k=1∑nak−k=1∑n−1ak=(kn)2−{k(n−1)}2

\hookrightarrow\displaystyle a_{n}=(kn)^{2}-\{k(n-1)\}^{2}\ (\text{for }n\geq2)↪an=(kn)2−{k(n−1)}2 (for n≥2)

\hookrightarrow a_{n}=k^{2}(n^{2}-n^{2}+2n-1)\ (\text{for }n\geq2)↪an=k2(n2−n2+2n−1) (for n≥2)

\hookrightarrow a_{n}=k^{2}(2n-1)\ (\text{for }n\geq2)↪an=k2(2n−1) (for n≥2)

a_{1}a1 follows the sequence.

\hookrightarrow a_{1}=k^{2}↪a1=k2

Choosing k=1k=1 we get,

\hookrightarrow a_{n}=2n-1\implies 1+3+5+7+\cdots+(2n-1)=n^{2}↪an=2n−1⟹1+3+5+7+⋯+(2n−1)=n2

Now, we found that adding the first n terms of odd numbers obtain n^{2}n2 . This also proves that the sum of the given sequence 4, 12, 20, …, 8n-4 obtain (2n)², it is a case where k=2k=2 .

$Solution:-</p><p>The sequence has a common difference of 8, and the first term 4.</p><p>The formula for series of an arithmetic progression is,</p><p>\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}=\dfrac{n\{2a+(n-1)d\}}{2}↪k=1∑nak=2n{2a+(n−1)d}</p><p>where aa is the first term, dd is the common difference, and nn is the term number.</p><p>\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}=\dfrac{n\{8+8(n-1)\}}{2}↪k=1∑nak=2n{8+8(n−1)}</p><p>\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}=4n^{2}↪k=1∑nak=4n2</p><p>\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}=(2n)^{2}↪k=1∑nak=(2n)2</p><p>So, the sum of the sequence is always a perfect square.</p><p>\large\text{\underline{Maths activity:-}}Maths activity:-</p><p>Also, we can make another perfect square sum.</p><p>Since,</p><p>\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}-\sum^{n-1}_{k=1}a_{k}=a_{n}\ (n\geq2)\text{ and }\sum^{1}_{k=1}a_{k}=a_{1}↪k=1∑nak−k=1∑n−1ak=an (n≥2) and k=1∑1ak=a1</p><p>Put a perfect square in terms of nn . We get,</p><p>\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}=(kn)^{2}↪k=1∑nak=(kn)2</p><p>Then,</p><p>\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}-\sum^{n-1}_{k=1}a_{k}=(kn)^{2}-\{k(n-1)\}^{2}↪k=1∑nak−k=1∑n−1ak=(kn)2−{k(n−1)}2</p><p>\hookrightarrow\displaystyle a_{n}=(kn)^{2}-\{k(n-1)\}^{2}\ (\text{for }n\geq2)↪an=(kn)2−{k(n−1)}2 (for n≥2)</p><p>\hookrightarrow a_{n}=k^{2}(n^{2}-n^{2}+2n-1)\ (\text{for }n\geq2)↪an=k2(n2−n2+2n−1) (for n≥2)</p><p>\hookrightarrow a_{n}=k^{2}(2n-1)\ (\text{for }n\geq2)↪an=k2(2n−1) (for n≥2)</p><p>a_{1}a1 follows the sequence.</p><p>\hookrightarrow a_{1}=k^{2}↪a1=k2</p><p>Choosing k=1k=1 we get,</p><p>\hookrightarrow a_{n}=2n-1\implies 1+3+5+7+\cdots+(2n-1)=n^{2}↪an=2n−1⟹1+3+5+7+⋯+(2n−1)=n2</p><p>Now, we found that adding the first n terms of odd numbers obtain n^{2}n2 . This also proves that the sum of the given sequence 4, 12, 20, …, 8n-4 obtain (2n)², it is a case where k=2k=2 .$