Question

JEE QUESTION SOLVE IT DUDE....​

Answer 1

First, make powers of 2 and those of 3 in the limit separately into two groups as follows:

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\left((2\times2^3\times2^5\times\dots\times2^{n-1})\times(3^2\times3^4\times3^6\times\dots\times3^n)\right)^{\frac{1}{n^2+1}}$

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\left(\left(2^{1+3+5+\,\dots\,+(n-1)}\right)\times\left(3^{2+4+6+\,\dots\,+n}\right)\right)^{\frac{1}{n^2+1}}$

We know,

• sum of first $n$ positive odd numbers is $n^2.$

• sum of first $n$ positive even numbers is $n^2+n$

Since each group contained $\dfrac{n}{2}$ terms,

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\left(2^{\left(\frac{n}{2}\right)^2}\times3^{\left(\frac{n}{2}\right)^2+\frac{n}{2}}\right)^{\frac{1}{n^2+1}}$

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\left(2^{\frac{n^2}{4}}\times3^{\frac{n^2+2n}{4}}\right)^{\frac{1}{n^2+1}}$

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\left[\left(2^{\frac{n^2}{4}}\right)^{\frac{1}{n^2+1}}\times\left(3^{\frac{n^2+2n}{4}}\right)^{\frac{1}{n^2+1}}\right]$

$\displaystyle\longrightarrow L=\lim_{n\to\infty}2^{\frac{n^2}{4(n^2+1)}}\times\lim_{n\to\infty}3^{\frac{n^2+2n}{4(n^2+1)}}$

$\longrightarrow L=2^{\displaystyle\lim_{n\to\infty}\frac{n^2}{4(n^2+1)}}\ \times\ 3^{\displaystyle\lim_{n\to\infty}\frac{n^2+2n}{4(n^2+1)}}\quad\quad\dots(1)$

Let us evaluate each limit.

$\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2}{4(n^2+1)}=\dfrac{1}{4}\lim_{n\to\infty}\dfrac{n^2}{n^2+1}$

Dividing both numerator and denominator by $\dfrac{1}{n^2},$

$\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2}{4(n^2+1)}=\dfrac{1}{4}\lim_{n\to\infty}\dfrac{\left(\frac{n^2}{n^2}\right)}{\left(\frac{n^2+1}{n^2}\right)}$

$\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2}{4(n^2+1)}=\dfrac{1}{4}\lim_{n\to\infty}\dfrac{1}{1+\frac{1}{n^2}}$

Since $\dfrac{1}{n^2}\to0$ as $n\to\infty,$

$\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2}{4(n^2+1)}=\dfrac{1}{4}\cdot\dfrac{1}{1+0}$

$\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2}{4(n^2+1)}=\dfrac{1}{4}$

And,

$\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2+2n}{4(n^2+1)}=\dfrac{1}{4}\lim_{n\to\infty}\dfrac{n^2+2n}{n^2+1}$

Dividing both numerator and denominator by $\dfrac{1}{n^2},$

$\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2+2n}{4(n^2+1)}=\dfrac{1}{4}\lim_{n\to\infty}\dfrac{\left(\frac{n^2+2n}{n^2}\right)}{\left(\frac{n^2+1}{n^2}\right)}$

$\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2+2n}{4(n^2+1)}=\dfrac{1}{4}\lim_{n\to\infty}\dfrac{1+\frac{2}{n}}{1+\frac{1}{n^2}}$

Since $\dfrac{1}{n^2}\to0$ as $n\to\infty,$

$\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2+2n}{4(n^2+1)}=\dfrac{1}{4}\cdot\dfrac{1+0}{1+0}$

$\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2+2n}{4(n^2+1)}=\dfrac{1}{4}$

Then (1) becomes,

$\longrightarrow L=2^{\frac{1}{4}}\times3^{\frac{1}{4}}$

$\longrightarrow L=6^{\frac{1}{4}}$

$\longrightarrow\underline{\underline{L^4=6}}}}$

Hence 6 is the answer.

Answer 2

Answer:

First, make powers of 2 and those of 3 in the limit separately into two groups as follows:

\displaystyle\longrightarrow L=\lim_{n\to\infty}\left((2\times2^3\times2^5\times\dots\times2^{n-1})\times(3^2\times3^4\times3^6\times\dots\times3^n)\right)^{\frac{1}{n^2+1}}⟶L=

n→∞

lim

((2×2

3

×2

5

×⋯×2

n−1

)×(3

2

×3

4

×3

6

×⋯×3

n

))

n

2

+1

1

\displaystyle\longrightarrow L=\lim_{n\to\infty}\left(\left(2^{1+3+5+\,\dots\,+(n-1)}\right)\times\left(3^{2+4+6+\,\dots\,+n}\right)\right)^{\frac{1}{n^2+1}}⟶L=

n→∞

lim

((2

1+3+5+…+(n−1)

)×(3

2+4+6+…+n

))

n

2

+1

1

We know,

sum of first nn positive odd numbers is n^2.n

2

.

sum of first nn positive even numbers is n^2+nn

2

+n

Since each group contained \dfrac{n}{2}

2

n

terms,

\displaystyle\longrightarrow L=\lim_{n\to\infty}\left(2^{\left(\frac{n}{2}\right)^2}\times3^{\left(\frac{n}{2}\right)^2+\frac{n}{2}}\right)^{\frac{1}{n^2+1}}⟶L=

n→∞

lim

(2

(

2

n

)

2

×3

(

2

n

)

2

+

2

n

)

n

2

+1

1

\displaystyle\longrightarrow L=\lim_{n\to\infty}\left(2^{\frac{n^2}{4}}\times3^{\frac{n^2+2n}{4}}\right)^{\frac{1}{n^2+1}}⟶L=

n→∞

lim

(2

4

n

2

×3

4

n

2

+2n

)

n

2

+1

1

\displaystyle\longrightarrow L=\lim_{n\to\infty}\left[\left(2^{\frac{n^2}{4}}\right)^{\frac{1}{n^2+1}}\times\left(3^{\frac{n^2+2n}{4}}\right)^{\frac{1}{n^2+1}}\right]⟶L=

n→∞

lim

[(2

4

n

2

)

n

2

+1

1

×(3

4

n

2

+2n

)

n

2

+1

1

]

\displaystyle\longrightarrow L=\lim_{n\to\infty}2^{\frac{n^2}{4(n^2+1)}}\times\lim_{n\to\infty}3^{\frac{n^2+2n}{4(n^2+1)}}⟶L=

n→∞

lim

2

4(n

2

+1)

n

2

×

n→∞

lim

3

4(n

2

+1)

n

2

+2n

\longrightarrow L=2^{\displaystyle\lim_{n\to\infty}\frac{n^2}{4(n^2+1)}}\ \times\ 3^{\displaystyle\lim_{n\to\infty}\frac{n^2+2n}{4(n^2+1)}}\quad\quad\dots(1)⟶L=2

n→∞

lim

4(n

2

+1)

n

2

× 3

n→∞

lim

4(n

2

+1)

n

2

+2n

…(1)

Let us evaluate each limit.

\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2}{4(n^2+1)}=\dfrac{1}{4}\lim_{n\to\infty}\dfrac{n^2}{n^2+1}⟶

n→∞

lim

4(n

2

+1)

n

2

=

4

1

n→∞

lim

n

2

+1

n

2

Dividing both numerator and denominator by \dfrac{1}{n^2},

n

2

1

,

\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2}{4(n^2+1)}=\dfrac{1}{4}\lim_{n\to\infty}\dfrac{\left(\frac{n^2}{n^2}\right)}{\left(\frac{n^2+1}{n^2}\right)}⟶

n→∞

lim

4(n

2

+1)

n

2

=

4

1

n→∞

lim

(

n

2

n

2

+1

)

(

n

2

n

2

)

\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2}{4(n^2+1)}=\dfrac{1}{4}\lim_{n\to\infty}\dfrac{1}{1+\frac{1}{n^2}}⟶

n→∞

lim

4(n

2

+1)

n

2

=

4

1

n→∞

lim

1+

n

2

1

1

Since \dfrac{1}{n^2}\to0

n

2

1

→0 as n\to\infty,n→∞,

\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2}{4(n^2+1)}=\dfrac{1}{4}\cdot\dfrac{1}{1+0}⟶

n→∞

lim

4(n

2

+1)

n

2

=

4

1

⋅

1+0

1

\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2}{4(n^2+1)}=\dfrac{1}{4}⟶

n→∞

lim

4(n

2

+1)

n

2

=

4

1

And,

\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2+2n}{4(n^2+1)}=\dfrac{1}{4}\lim_{n\to\infty}\dfrac{n^2+2n}{n^2+1}⟶

n→∞

lim

4(n

2

+1)

n

2

+2n

=

4

1

n→∞

lim

n

2

+1

n

2

+2n

Dividing both numerator and denominator by \dfrac{1}{n^2},

n

2

1

,

\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2+2n}{4(n^2+1)}=\dfrac{1}{4}\lim_{n\to\infty}\dfrac{\left(\frac{n^2+2n}{n^2}\right)}{\left(\frac{n^2+1}{n^2}\right)}⟶

n→∞

lim

4(n

2

+1)

n

2

+2n

=

4

1

n→∞

lim

(

n

2

n

2

+1

)

(

n

2

n

2

+2n

)

\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2+2n}{4(n^2+1)}=\dfrac{1}{4}\lim_{n\to\infty}\dfrac{1+\frac{2}{n}}{1+\frac{1}{n^2}}⟶

n→∞

lim

4(n

2

+1)

n

2

+2n

=

4

1

n→∞

lim

1+

n

2

1

1+

n

2

Since \dfrac{1}{n^2}\to0

n

2

1

→0 as n\to\infty,n→∞,

\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2+2n}{4(n^2+1)}=\dfrac{1}{4}\cdot\dfrac{1+0}{1+0}⟶

n→∞

lim

4(n

2

+1)

n

2

+2n

=

4

1

⋅

1+0

1+0

1

→0 as n\to\infty,n→∞,

\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2+2n}{4(n^2+1)}=\dfrac{1}{4}\cdot\dfrac{1+0}{1+0}⟶

n→∞

lim

4(n

2

+1)

n

2

+2n

=

4

1

⋅

1+0

1+0

\displaystyle\longrightarrow\lim_{n\to\infty}\dfrac{n^2+2n}{4(n^2+1)}=\dfrac{1}{4}⟶

n→∞

lim

4(n

2

+1)

n

2

+2n

=

4

1

Then (1) becomes,

\longrightarrow L=2^{\frac{1}{4}}\times3^{\frac{1}{4}}⟶L=2

4

1

×3

4

1

\longrightarrow L=6^{\frac{1}{4}}⟶L=6

4

1

$$\longrightarrow\underline{\underline{L^4=6}}}}$$

Hence 6 is the answer.