MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Middle Terms

If n is even ...

Question

If $n$ is even and the middle term in the expansion of ${(x^{2} + \frac{1}{x})}^{n}$ is $924 x^{6}$ , then $n$ is equal to

A

10

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

12

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C

14

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $12$
$∵ n$ is even
$∴$ The middle term $= {(\frac{n}{2} + 1)}^{th}$ term

$\Rightarrow T_{(n / 2 + 1)} =^{n} C_{n / 2} (x^{2})^{n / 2} {(\frac{1}{x})}^{n / 2}$

$= 924 x^{6}$ (given)

$\Rightarrow x^{n / 2} = x^{6} \Rightarrow n = 12$

Suggest Corrections

thumbs-up

1

Similar questions

Q. The middle term in the expansion of

{(x^{2} + \frac{1}{x})}^{n}

924 x^{6}

n

n

Q. If the middle term in the expansion of

{(x^{2} + \frac{1}{x})}^{n}

924 x^{6}

n =

Q. Find the coefficient of the middle term in the expansion of

(1 + x)^{n}

when n is even and the coefficient of middle terms if n is odd. Your answer must not be in factorial notations.

{(x^{2} + \frac{1}{x})}^{n}

has exactly one middle term which is equal to

α . x^{3}

then the value of

(α + n)

n \in N

)

Q. Show that the middle term in the expansion of

(1 + x)^{n}

6 x^{2}

n = 4

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon