Question

If (r + 1)ᵗʰ term is the first negative term in the expansion
of (1 + x)⁷/², then the value of r, is
(a) 5 (b) 6 (c) 4 (d) 7

Answer 1

Answer:

c is the correct answer plz mark it as brainliest and follow me

Answer 2

(a) 5

Explanation:

As we know that $\frac{7}{2}$ is a rational number. Hence, for n to a rational number, the expansion is,

$(1+x)^n=1+nx+\frac{n(n-1)x^2}{2!}+ \frac{n(n-1)(n-2)x^3}{3!}+.....$

So, for $n=\frac{7}{2}$, the expansion will be,

$(1+x)^\frac{7}{2} =1+\frac{7}{2}x+\frac{\frac{7}{2}(\frac{7}{2}-1)x^2}{2!}+ \frac{\frac{7}{2}(\frac{7}{2}-1)(\frac{7}{2}-2)x^3}{3!}+.....$

Hence $T_{r+1}$ term will be,

$\frac{\frac{7}{2}(\frac{7}{2}-1)(\frac{7}{2}-2).....(\frac{7}{2}-r+1)}{r!}$

As per given information in the question if  $T_{r+1}$ s the first negative term , then

$\frac{7}{2}-r+1 < 0\\ \\\implies\frac{9}{2}-r < 0\\\\\implies r > \frac{9}{2}$

Hence, the possible value of r is 5.

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