Question

If $\frac{1}{\sqrt{4x+1}}[{\left(\frac{1+\sqrt{4x+1}}{2}\right)}^n-{\left(\frac{1-\sqrt{4x+1}}{2}\right)}^n]=a_0+a_{1}x+a_2{x}^2+....+a_5{x}^5$

Find $n$.

Answer 1

$\frac{1}{\sqrt{4x+1}}[{\left(\frac{1+\sqrt{4x+1}}{2}\right)}^n-{\left(\frac{1-\sqrt{4x+1}}{2}\right)}^n]$ has the highest power of $x$ to be $5$.
Let $\sqrt{4x+1}=y$

And we know that ${(1+y)}^n-{(1-y)}^n=2{(}^n{C}_{1}z{+}^n{C}_3{z}^3+.........{.}^n{C}_n{z}^n)$

So applying this we get $\frac{2}{y\times 2^n}{(}^n{C}_{1}y{+}^n{C}_3{y}^3+.........{.}^n{C}_n{y}^n)$

So highest power is $y^{n-1}$ that is ${(4x+1)}^{\frac{n-1}{2}}$ So highest power of $x$ in the given expansion is $\frac{n-1}{2}=5\Rightarrow n=11$