Question

If $x$ is positive, the first negative term in the expansion of ${(1+x)}^{\frac{27}{5}}$ is

• A. $5^{th}$ term
• B. $7^{th}$ term
• C. $6^{th}$ term
• D. $8^{th}$ term

Answer 1

Answer: (B)

$7^{th}$ term

$(1+x{)}^{\frac{9}{2}}$$=(1+x{)}^{4+\frac{1}{2}}$
$=(1+x{)}^4(1+x{)}^{\frac{1}{2}}$
$=(1+4x+6x^2+4x^3+x^4)(1+x{)}^{\frac{1}{2}}$
$=(1+4x+6x^2+4x^3+x^4)(1+\frac{x}{2}-\frac{x^2}{8}...\infty )$
Hence the first negative term will be the $7th$ term.