Question

Let $a=3^{1/203}+1$ and $f\left(n\right)={}^n{C}_0{a}^{n-1}-{}^n{C}_1{a}^{n-2}+{}^n{C}_2{a}^{n-3}-\cdots +(-1{)}^{n-1}{}^n{C}_{n-1}{a}^0$ where $n\ge 3$. If $f\left(2030\right)+f\left(2031\right)=3^x(\frac{y}{a}-1)$, then the value of $\frac{x}{y}$ is

• A. $6$
• B. $5$
• C. $4$
• D. $4.5$

Answer 1

The correct option is B $5$

$f\left(n\right)={}^n{C}_0{a}^{n-1}-{}^n{C}_1{a}^{n-2}+{}^n{C}_2{a}^{n-3}-\cdots +(-1{)}^{n-1}{}^n{C}_{n-1}{a}^0\Rightarrow af(n)={}^n{C}_0{a}^n-{}^n{C}_1{a}^{n-1}+{}^n{C}_2{a}^{n-2}-\cdots +(-1{)}^{n-1}{}^n{C}_{n-1}a+(-1{)}^n{}^n{C}_n{a}^0-(-1{)}^n{}^n{C}_n{a}^0\Rightarrow af\left(n\right)=(1-a{)}^n-(-1{)}^n\Rightarrow af\left(2030\right)=(1-a{)}^{2030}-1\Rightarrow af(2031)=(1-a{)}^{2031}+1\Rightarrow a\left(f\right(2030)+f(2031\left)\right)=(1-a{)}^{2030}+(1-a{)}^{2031}$

We know that,
$a=3^{1/203}+1\Rightarrow 1-a=-\left(3^{1/203}\right)$
$a\left(f\right(2030)+f(2031\left)\right)=(1-a{)}^{2030}(1+(1-a))\Rightarrow a(f\left(2030\right)+f\left(2031\right))=3^{10}(2-a)\Rightarrow f(2030)+f(2031)=3^{10}\times \frac{(2-a)}{a}\Rightarrow f(2030)+f(2031)=3^{10}\times (\frac{2}{a}-1)\Rightarrow 3^x(\frac{y}{a}-1)=3^{10}\times (\frac{2}{a}-1)\Rightarrow x=10,y=2\Rightarrow \frac{x}{y}=5$