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If  x is positive, the first negative term in the expansion of (1 + x)27/5 is [2003]
  • a)
    6th term
  • b)
    7th term
  • c)
    5th term
  • d)
    8th term.
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
If x is positive, the first negative term in the expansion of (1 + x)2...
For first negative term,  n - r + 1 < 0 ⇒ r >n+1
Therefore, first negative term  is T8 .
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Most Upvoted Answer
If x is positive, the first negative term in the expansion of (1 + x)2...
Solution:

Given, (1+x)^(27/5)

Using binomial theorem, we can write the expansion of the given expression as:

(1+x)^(27/5) = C(0,27/5) + C(1,27/5)x + C(2,27/5)x^2 + .... + C(r,27/5)x^r + ....

where C(r,27/5) = 27/5C(r-1,22/5)/r

We are looking for the first negative term in the expansion of (1+x)^(27/5)

To find the first negative term, we need to find the smallest value of r such that C(r,27/5)x^r < />

Since x is positive, we can ignore the sign of C(r,27/5) and look for the smallest value of r such that C(r,27/5) < />

C(r,27/5) < 0="" if="" and="" only="" if="" r="" /> 8

Therefore, the first negative term in the expansion of (1+x)^(27/5) is the 8th term, i.e., C(8,27/5)x^8

Hence, the correct answer is option D, i.e., the 8th term.
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If x is positive, the first negative term in the expansion of (1 + x)27/5 is [2003]a)6th termb)7th termc)5th termd)8th term.Correct answer is option 'D'. Can you explain this answer?
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