Question

If ${\left(x+\frac{1}{x}+1\right)}^6=a_0+\left(a_{1}x+\frac{b_1}{x}\right)+\left(a_2{x}^2+\frac{b_2}{x^2}\right)+\dots \dots +\left(a_6{x}^6+\frac{b_6}{x^6}\right)$, then

• A. $a_0=141$
• B. $a_5=6$
• C. $\sum _{i=1}^6\left(a_i+b_i\right)=588$
• D. $\sum _{i=1}^6\left(a_i+b_i\right)=3^6$

Answer 1

Answer: (A), (B), (C)

${\left(x+\frac{1}{x}+1\right)}^6=\sum _{r=1}^6{}^6{C}_r{\left(x+\frac{1}{x}\right)}^r$
(A) $a_0=1+{}^6{C}_2\cdot {}^2{C}_1+{}^6{C}_4\cdot {}^4{C}_2+{}^6{C}_6\cdot {}^6{C}_3=1+30+90+20=141$.
(B) $a_5={}^6{C}_5\cdot$ coefficient of $x^5$ in ${\left(x+\frac{1}{x}\right)}^5={}^6{C}_5\cdot {}^5{C}_0=6$
(C) Putting $x=1$
$a_0+\left(a_1+b_1\right)+\left(a_2+b_2\right)+\dots \dots +\left(a_6+b_6\right)=3^6$
$\therefore \sum _{i=1}^6\left(a_i+b_i\right)=3^6-141=588$