The cofficient of middle term in the expansion of ${(1 + x)}^{10}$

\frac{10!}{5! 6!}

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

\frac{10!}{{(5!)}^{2}}

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

\frac{10!}{5! 7!}

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

N o n e o f t h e s e

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

Open in App

Solution

The correct option is B $\frac{10!}{{(5!)}^{2}}$
Expansion will contain 11 terms so middle term will be the 6th term
which will be $=^{10} C_{5} x^{5}$

So, its cofficient $=^{10} C_{5} = \frac{10!}{5! 5!}$

Suggest Corrections

Similar questions

The middle term in the expansion of (x + y)¹⁰ is the

{(1 + 7 x)}^{10}

^{10} C_{5} 7^{5}

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

The coefficient of the middle term in the binomial expansion of $(1 + a x)^{4}$ and of $(1 - a x)^{6}$ is the same, if a equals:

x

(1 + α x)^{4}

(1 - α x)^{6}

α

Join BYJU'S Learning Program