Question

If x is positive, the first negative term in the expansion \(\rm (1+x)^{\frac{9}{2}}\) is 
1. 5th term 
2. 6th term 
3. 7th term 
4. None of the above

Answer 1

Correct Answer - Option 3 : 7th term 

Concept:

The general term in the expansion of (a + b)ⁿ is given by, \(\rm T_{r+1}=^nC_r(a)^{n-r}(b)^r\), where, r cannot be fractional.

 \(\rm ^nC_r=\frac{n\times (n-1)\times...\times(n-r+1)}{r!}\)

 

Calculation:

Given binomial expansion: \(\rm (1+x)^{\frac{9}{2}}\)

\(\rm T_{r+1}=^{\frac{9}{2}}C_r(1)^{n-r}(x)^r\)

= \(\rm \frac{\frac92\times(\frac92-1)\times....(\frac92-r+1) }{r!}\times (x)^r\)

Now, for this term to become negative, 

\(\rm (\frac92-r+1) \) < 0

⇒r > 11/2

⇒r > 5.5

⇒r = 6       (∵ r cannot be fractional)

⇒ r + 1 = 6 + 1 = 7

So, first negative term is 7th term

Hence, option (3) is correct.