When |x|>3, then coefficient of \;1/xⁿ in the expansion of x³/2}(3+x)¹/2} is

Correct Answer is : (−1)n+11⋅3⋅5⋯(2n+1)2n+2(n+2)!3n+2(-1)ⁿ+1} {1 · 3 · 5 ·s (2n+1)}{2ⁿ+2}(n+2)!} 3ⁿ+2}(−1)n+12n+2(n+2)!1⋅3⋅5⋯(2n+1)3n+2

(-1)ⁿ 1 · 3 · 5 ·s (2n-1)/2ⁿ n! 3ⁿ

(-1)ⁿ+1} {1 · 3 · 5 ·s (2n+1)}{2ⁿ+2}(n+2)!} 3ⁿ+2}

(-1)ⁿ+1} 1 · 3 · 5 ·s (2n-1)/2ⁿ n! 3ⁿ+1}

(-1)ⁿ+1} {1 · 3 · 5 ·s (2n+1)}{2ⁿ+3}(n+2)!} 3ⁿ+1}

Correct Answer is : (−1)n+11⋅3⋅5⋯(2n+1)2n+2(n+2)!3n+2(-1)ⁿ+1} {1 · 3 · 5 ·s (2n+1)}{2ⁿ+2}(n+2)!} 3ⁿ+2}(−1)n+12n+2(n+2)!1⋅3⋅5⋯(2n+1)3n+2

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TG EAMCET 2 May 2025 Shift 2 Paper

Question : 20 of 160

Marks: +1, -0

When

|x|>3

\;\frac{1}{x^n}

x^{3/2}(3+x)^{1/2}

Solution:

We have,

x^{3/2}(3+x)^{1/2}

\; = x^{3/2} x^{1/2} \left(1+\; \frac{3}{x}\right)^{1/2}

\; = x^2 \left(1+\; \frac{3}{x}\right)^{1/2}

Now, coefficient of general term of

\left(1+\; \frac{3}{x}\right)^{1/2}

\; \frac{ \left(\frac{1}{2}\right) \left(-\frac{1}{2}\right) \left(-\frac{1}{3}\right) \cdots \left(\frac{1}{2}-k+1\right) }{k!}

\; \frac{(-1)^{k+1} 1 \cdot 3 \cdot 5 \cdots (2k-3)}{2k k!}

On multiply by

x^2

, then

x^2 \left(1+\; \frac{3}{x}\right)^{1/2} = \sum\limits_{k=0}^{\infty} {}^{1/2}C_k 3^k x^{2-k}

For coefficient of

\; \frac{1}{x^n}, 2-k=-n

\Rightarrow \;\; k=n+2

Coefficient of

\; \frac{1}{x^n}

\;\; \frac{(-1)^{n+3} 1 \cdot 3 \cdot 5 \cdots (2(n+2)-3) 3^{n+2}}{2^{n+2}(n+2)!}

\; \;\; = \frac{(-1)^{n+1} 1 \cdot 3 \cdot 5 \cdots (2n+1)}{2^{n+1}(n+2)!} 3^{n+2}

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