Question

Find the coefficient of $x^6$ in the expansion of ${(1-2x)}^{\frac{-5}{2}}$.

• A. $\frac{1500}{16}$
• B. $\frac{15012}{16}$
• C. $\frac{15015}{16}$
• D. $\frac{15016}{16}$

Answer 1

The correct option is C $\frac{15015}{16}$

We know, ${(1+x)}^n=\frac{n\cdot (n-1)\cdot (n-2)...\cdot (n-r+1)}{r!}{x}^r$

So, ${(1-2x)}^n=\frac{n\cdot (n-1)\cdot (n-2)...\cdot (n-r+1)}{r!}{(-2)}^r{x}^r$

Hence, putting $r=6$ and $n=\frac{-5}{2},$ we get,

${(1-2x)}^{\frac{-5}{2}}=\frac{\frac{-5}{2}\cdot (\frac{-5}{2}-1)\cdot (\frac{-5}{2}-2)\cdot (\frac{-5}{2}-3)\cdot (\frac{-5}{2}-4)\cdot (\frac{-5}{2}-5)}{6!}{(-2)}^6{x}^6=\frac{15015}{16}$