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Q.
If $\left(x+\frac{1}{x}+1\right)^6=a_0+\left(a_1 x+\frac{b_1}{x}\right)+\left(a_2 x^2+\frac{b_2}{x^2}\right)+\ldots \ldots+\left(a_6 x^6+\frac{b_6}{x^6}\right)$, then
Binomial Theorem
Solution:
$\left(x+\frac{1}{x}+1\right)^6=\displaystyle\sum_{r=1}^6{ }^6 C_r\left(x+\frac{1}{x}\right)^r$
(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)+\ldots \ldots+\left(a_6+b_6\right)=3^6 $
$\therefore \displaystyle\sum_{i=1}^6\left(a_i+b_i\right)=3^6-141=588$