The middle term in the expansion of 1-x2 n is-A- 1 - 3 -2 n-1-n - xnB- None of theseC- 1 - 2 - 3 -2 n-1-n - 2n xnD- 1 - 3 - 5 -2 n-1-n - 2n xn

The correct option is D Middle term = T_n+1 = nbsp;^2n C nbsp;_n = nbsp; 1.2.3.......(2n 1)2^n/n!x^n = nbsp; 1.3.5......(2n 1)/n! nbsp;× nbsp; 2^n ...

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

Byju's Answer

Standard XI

Mathematics

Middle Terms

The middle te...

Question

The middle term in the expansion of $(1 + x)^{2 n}$ is:

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

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

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

Open in App

Solution

The correct option is B

Middle term

= $T_{n + 1}$ = $^{2 n} C_{n}$ = $\frac{1.2.3....... (2 n - 1) 2^{n}}{n!}$ $x^{n}$

= $\frac{1.3.5...... (2 n - 1)}{n!} \times \frac{2^{n} (1.2.3.......... n)}{n!} x^{n}$

= $\frac{1.3.5......... (2 n - 1)}{n!}$ $2^{n} x^{n}$ .

Suggest Corrections

Similar questions

The middle term in the expansion of $(1 + x)^{2 n}$ is

Q. The middle term in the expansion of

(1 + x)^{2 n}

Q. Shew that the middle term in the expansion of

(1 + x)^{2 n}

\frac{1.3.5... (2 n - 1)}{(n)!} 2^{n} x^{n}

Q. Show that the middle term in the expansion of

(1 + x)^{2 n}

\frac{1.3.5... (2 n - 1)}{n!}

2^{n} x^{n}

, where n is a positive integer.

Q. Show that the middle term in the expansion of

(1 + x)^{2 n}

\frac{1.3.5..... (2 n - 1)}{n!}

2^{n} x^{n}

; where n is a positive integer.

Join BYJU'S Learning Program

The middle term in the expansion of (1+x)2n is:

The correct option is B Middle term = Tn+1 = 2nCn = 1.2.3.......(2n−1)2nn!xn = 1.3.5......(2n−1)n!×2n(1.2.3..........n)n!xn = 1.3.5.........(2n−1)n! 2nxn.

The middle term in the expansion of $(1 + x)^{2 n}$ is:

The correct option is B

Middle term

= $T_{n + 1}$ = $^{2 n} C_{n}$ = $\frac{1.2.3....... (2 n - 1) 2^{n}}{n!}$ $x^{n}$

= $\frac{1.3.5...... (2 n - 1)}{n!} \times \frac{2^{n} (1.2.3.......... n)}{n!} x^{n}$

= $\frac{1.3.5......... (2 n - 1)}{n!}$ $2^{n} x^{n}$ .