Question

The middle term in the expansion of
${(2x + \dfrac{1}{x} )}^{12}$
is ..?​

Answer 1

$\huge\mathcal{\mid{\mid{\underline{\pink{Good\: Evening\:}}}{\mid{\mid}}}}$

Qᴜᴇsᴛɪᴏɴ ;-

The middle term in the expansion of

$\sf {(2x + \dfrac{1}{x} )}^{12}$

is _ .

$\huge{\orange{\boxed{\fcolorbox{lime}{aqua}{\pink{ANSWER}}}}}$

$= > \: n \: is \: even, \: so \: (\frac{n}{2} + 1)^{th} \: term \: . \\ \\ = > ( \frac{12}{2} + 1) \\ \\ = > {7}^{th} \: term$

We know that,

$\longmapsto~\sf\pink{T_{r\:+\:1}~=~^{n}C_{r}\:x^{n\:-\:r}\:y^r\:}$

Where,

• r = 7

• n = 12

• x = 2x

• y = 1/x

$\longmapsto~\rm{T_{7\:+\:1}~=~^{12}C_{7}\:(2x)^{12\:-\:7}\:(\dfrac{1}{x})^7\:}$

$\longmapsto~\rm{T_{6}~=~^{12}C_{7}\:(2x)^{5}\:\dfrac{1}{x^7}\:}$

$\longmapsto~\rm{T_{6}~=~^{12}C_{7}\times{2^5}\times{x^{5}}\times{x^{-7}}\:}$

$\longmapsto~\rm{T_{6}~=~792\times{32}\times{x^{-2}}\:}$

$\longmapsto~\rm\green{T_{6}~=~25344~{x^{-2}}\:}$

Answer 2

QUESTION

The middle term in the expansion of

${(2x + \dfrac{1}{x} )}^{12}$

is ..?

ANSWER

=>niseven,so(

2

n

+1)

th

term.

=>(

2

12

+1)

=>7

th

term

We know that,

\begin{gathered}\longmapsto~\sf\pink{T_{r\:+\:1}~=~^{n}C_{r}\:x^{n\:-\:r}\:y^r\:} \\ \end{gathered}

⟼ T

r+1

=

n

C

r

x

n−r

y

r

Where,

r = 7

n = 12

x = 2x

y = 1/x

\begin{gathered}\longmapsto~\rm{T_{7\:+\:1}~=~^{12}C_{7}\:(2x)^{12\:-\:7}\:(\dfrac{1}{x})^7\:} \\ \end{gathered}

⟼ T

7+1

=

12

C

7

(2x)

12−7

(

x

1

)

7

\begin{gathered}\longmapsto~\rm{T_{6}~=~^{12}C_{7}\:(2x)^{5}\:\dfrac{1}{x^7}\:} \\ \end{gathered}

⟼ T

6

=

12

C

7

(2x)

5

x

7

1

\begin{gathered}\longmapsto~\rm{T_{6}~=~^{12}C_{7}\times{2^5}\times{x^{5}}\times{x^{-7}}\:} \\ \end{gathered}

⟼ T

6

=

12

C

7

×2

5

×x

5

×x

−7

\begin{gathered}\longmapsto~\rm{T_{6}~=~792\times{32}\times{x^{-2}}\:} \\ \end{gathered}

⟼ T

6

= 792×32×x

−2

\begin{gathered}\longmapsto~\rm\green{T_{6}~=~25344~{x^{-2}}\:} \\ \end{gathered}

⟼ T

6

= 25344