Question

If $a=3^{1/223}+1$ and $f\left(n\right)=\sum _{r=1}^n(-1{)}^{r-1}{}^n{C}_{r-1}\cdot a^{n-r}$ for all $n\ge 3$, then the value of $f\left(2007\right)+f\left(2008\right)$ is

• A. $3^6$
• B. $3^9$
• C. $3^{12}$
• D. $3^{15}$

Answer 1

The correct option is B $3^9$

$f\left(n\right)=\sum _{r=1}^n(-1{)}^{r-1}{}^n{C}_{r-1}\cdot a^{n-r}=\frac{1}{a}\sum _{r=1}^n(-1{)}^{r-1}{}^n{C}_{r-1}\cdot a^{n-r+1}=\frac{1}{a}[{}^n{C}_0{a}^n-{}^n{C}_1{a}^{n-1}+\cdots +(-1{)}^{n-1}{}^n{C}_{n-1}{a}^1]=\frac{1}{a}\left[\right(a-1{)}^n-(-1{)}^n{C}_n]=\frac{1}{a}[3^{n/223}-(-1{)}^n]\Rightarrow f\left(n\right)=\frac{3^{n/223}-(-1{)}^n}{(3^{1/223}+1)}$
Now,
$f\left(2007\right)+f\left(2008\right)=\frac{3^{2007/223}+3^{2008/223}}{3^{1/223}+1}=\frac{3^9+3^{2008/223}}{3^{1/223}+1}=3^9\left[\frac{1+3^{1/223}}{3^{1/223}+1}\right]=3^9$