Question

If $\frac{1}{\sqrt{4x+1}}\{{\left(\frac{1+\sqrt{4x+1}}{2}\right)}^n-{\left(\frac{1-\sqrt{4x+1}}{2}\right)}^n\}=a_0+a_{1}x+....a_5{x}^5$, then $n=$

• A. $11$
• B. $9$
• C. $10$
• D. $Noneofthese$

Answer 1

The correct option is A $11$

$\frac{1}{\sqrt{4x+1}}\{{\left(\frac{1+\sqrt{4x+1}}{2}\right)}^n-{\left(\frac{1-\sqrt{4x+1}}{2}\right)}^n\}$

$\Rightarrow \frac{1}{\sqrt{4x+1}}\{{\left(\frac{1+\sqrt{4x+1}}{2}\right)}^{n/2}-{\left(\frac{1-\sqrt{4x+1}}{2}\right)}^{n/2}\}-[{\left(\frac{1+\sqrt{4x+1}}{2}\right)}^{n/2}-{\left(\frac{1-\sqrt{4x+1}}{2}\right)}^{n/2}]$

$\Rightarrow \frac{(x{)}^{n/2}}{(x{)}^{1/2}}=5$

$\Rightarrow (x{)}^{n/2-\frac{1}{2}}=5$

$\frac{n}{2}-\frac{1}{2}=5\Rightarrow \frac{n}{2}=\frac{11}{2}\Rightarrow n=11$