Question

If ${(x+\frac{1}{x}+1)}^6=a_0+(a_{1}x+\frac{b_1}{x})+(a_2{x}^2+\frac{b_2}{x^2})+...+(a_6{x}^6+\frac{b_6}{x^6})$ then find the value of $a_0$

Answer 1

$\because {(x+\frac{1}{x}+1)}^6={\sum }_{r=0}^{r=6}{}^6{C}_r{(x+\frac{1}{x})}^r$ for constant term $r$ must be even integer.

$\therefore a_0{=}^6{C}_0{+}^6{C}_2{\times }^6{C}_1{+}^6{C}_4{\times }^6{C}_2{+}^6{C}_6{\times }^6{C}_3$

$=1+\frac{6\times 5\times 4!}{4!\times 2}\times \frac{6\times 5!}{5!\times 1}+\frac{6\times 5\times 4!}{4!\times 2}\times \frac{6\times 5\times 4!}{4!\times 2}+1\times \frac{6\times 5\times 4\times 3!}{3!3!}$

$=1+30+90+20=141$