Question

Let $f\left(x\right)=\sum _{r=0}^{20}{a}_r{x}^r$ and $g\left(x\right)=\sum _{r=0}^9{b}_r{x}^r+\sum _{r=10}^{20}{x}^r.$ If $f\left(x\right)=g(x+1),$ then the value of $a_{10}$ is

• A. ${}^{21}{C}_{11}$
• B. ${}^{19}{C}_9$
• C. ${}^{19}{C}_{11}$
• D. ${}^{21}{C}_9$

Answer 1

The correct option is A ${}^{21}{C}_{11}$

Given $f\left(x\right)=\sum _{r=0}^{20}{a}_r{x}^r$ and $g\left(x\right)=\sum _{r=0}^9{b}_r{x}^r+\sum _{r=10}^{20}{x}^r$

According to given condition,
$f\left(x\right)=g(x+1)\Rightarrow \sum _{r=0}^{20}{a}_r{x}^r=\sum _{r=0}^9{b}_r(x+1{)}^r+\sum _{r=10}^{20}(x+1{)}^r\Rightarrow \sum _{r=0}^{20}{a}_r{x}^r=\sum _{r=0}^9{b}_r(x+1{)}^r+(x+1{)}^{10}\left\{\frac{(x+1{)}^{11}-1}{x+1-1}\right\}$
$\Rightarrow \sum _{r=0}^{20}{a}_r{x}^r=\sum _{r=0}^9{b}_r(x+1{)}^r+\frac{(x+1{)}^{21}-(x+1{)}^{10}}{x}$

Comparing the coefficients of $x^{10}$, we get
$a_{10}=$ coefficient of $x^{11}$ in $\left[\right(x+1{)}^{21}-(x+1{)}^{10}]$
$\Rightarrow a_{10}=$ coefficient of $x^{11}$ in $(x+1{)}^{21}$
$\therefore a_{10}={}^{21}{C}_{11}$