Question

The coefficient of $(r+1)th$ term of ${(x+\frac{1}{x})}^{20}$ when expanded in the descending powers of $x$ is equal to the coefficient of the $6th$ term of ${(x^2+2+\frac{1}{x^2})}^{10}$ when expanded in ascending powers of $x$. The value of $r$ is

• A. $5$
• B. $6$
• C. $14$
• D. $15$

Answer 1

Answer: (A), (D)

$5$
$15$

General term in the expansion of ${(x+\frac{1}{x})}^{20}$ is given by:
$T_{r+1}={}^{20}{C}_r{x}^{20-r}{x}^{-r}$
$T_{r+1}={}^{20}{C}_r{x}^{20-2r}$ ...(i)
Now ${(x^2+2+\frac{1}{x^2})}^{10}$ can be written as
${\left[{(x+\frac{1}{x})}^2\right]}^{10}$
$={(x+\frac{1}{x})}^{20}$
Therefore
$r+1=6$
$r=5$ ....(since both the expansions are identical).
Hence the corresponding coefficients, will be
${}^{20}{C}_5$
$={}^{20}{C}_{20-5}$
$={}^{20}{C}_{15}$