Statement -1 : The term independent of x in the expansion of ${(x^{2} + \frac{1}{x^{2}} + 2)}^{25} i s^{50} C_{25} .$
Statement-2 : In a binomial expansion middle term is independent of x.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

Statement -1 is true, Statement-2 is true; Statement-2 is correct explanation for Statement -1

Statement -1 is true, Statement-2 is true; Statement-2 is not correct explanation for Statement -1

Statement -1 is false, Statement -2 is true

Statement -1 is true, Statement -2 is false

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

We have, ${(x^{2} + \frac{1}{x^{2}} + 2)}^{25} = {(x + \frac{1}{x})}^{50} .$

$∴ T_{r + 1} =^{50} C_{r} . x^{50 - r} . {(\frac{1}{x})}^{r} =^{50} C_{r} x^{50 - 2 r}$

For independent of x, we put $50 - 2 r = 0 \Rightarrow r = 25$

$∴ T_{25 + 1} =^{50} C_{25}$

But in binomial expansion of ${(x + a)}^{n},$ middle terms is independent of x, iff $x . a = 1.$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE