Question

The degree of the polynomial $\frac{1}{\sqrt{4x+1}}[{\left(\frac{1+\sqrt{4x+1}}{2}\right)}^7-{\left(\frac{1-\sqrt{4x+1}}{2}\right)}^7]$ is

Answer 1

Given
$\frac{1}{\sqrt{4x+1}}[{\left(\frac{1+\sqrt{4x+1}}{2}\right)}^7-{\left(\frac{1-\sqrt{4x+1}}{2}\right)}^7]=\frac{1}{2^7\sqrt{4x+1}}\left[\right(1+\sqrt{4x+1}{)}^7-(1-\sqrt{4x+1}{)}^7]$

We know that,
$(1+x{)}^7-(1-x{)}^7=2[{}^7{C}_{1}x+{}^7{C}_3{x}^3+\dots +{}^7{C}_7{x}^7]$
Putting $x=\sqrt{4x+1}$, we get
$\frac{1}{\sqrt{4x+1}}[{\left(\frac{1+\sqrt{4x+1}}{2}\right)}^7-{\left(\frac{1-\sqrt{4x+1}}{2}\right)}^7]=\frac{2}{2^7\sqrt{4x+1}}\left[{}^7{C}_1\right(\sqrt{4x+1})+{}^7{C}_3(\sqrt{4x+1}{)}^3+\dots +{}^7{C}_7\left(\sqrt{4x+1}{)}^7\right]=\frac{1}{2^6}[{}^7{C}_1+{}^7{C}_3(\sqrt{4x+1}{)}^2+\dots +{}^7{C}_7\left(\sqrt{4x+1}{)}^6\right]=\frac{1}{2^6}[{}^7{C}_1+{}^7{C}_3(4x+1)+\dots +{}^7{C}_7(4x+1{)}^3]$

Hence, the degree of the polynomial is $3$