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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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.