Question

If $\frac{1}{\sqrt{4x+1}}[{\left(\frac{1+\sqrt{4x+1}}{2}\right)}^n-{\left(\frac{1-\sqrt{4x+1}}{2}\right)}^n]$ = $a_0+a_{1}x+...+a_5{x}^5$ show that n = 11

Answer 1

Let $\sqrt{4x+1}=y\therefore 4x+1=y^2$
or $x^5={\left(\frac{y^2-1}{4}\right)}^5$ i.e., the power of y is 10
$E=\frac{1}{y}[{\left(\frac{1+y}{2}\right)}^n-{\left(\frac{1-y}{2}\right)}^n]$
$\frac{1}{y}\cdot \frac{1}{2^n}[2{(}^n{C}_{1}y+{}^n{C}_3{y}^3+...+{}^n{C}_n{y}^n)]$
For $x^5$, E should contain $y^{1}0$ or the numerator of E should have the term of $y^{11}$.
Hence $n=11$