Question

Let $a=3^{\frac{1}{223}}+1$ and let $f\left(n\right){=}^n{C}_0.a^{n-1}{-}^n{C}_1.a^{n-2}{+}^n{C}_2.a^{n-3}-....+(-1{)}^{n-1}{.}^n{C}_{n-1}.a^0$. If the values of $f\left(2007\right)+f\left(2008\right)=9k$ where $kϵN$, then find $k$.

• A. $2187$
• B. $2324$
• C. $2543$
• D. $2874$

Answer 1

Answer: (A)

$2187$

$f\left(n\right)=\left(a^{n-1}\right)[{}^n{C}_0{-}^n{C}_1\left(a^{-1}\right){+}^n{C}_2{\left(a^{-1}\right)}^2+.......{.}^n{C}_{n-2}{\left(a^{-1}\right)}^{n-2}{+}^n{C}_{n-1}{\left(a^{n-1}\right)}^{n-1}]$

$f\left(n\right)=\left(a^{n-1}\right)\{\left[{(1-\frac{1}{a})}^n\right]-{(-1)}^n.\left(a^{n-1}\right)\}$

$=\left(a^{n-1}\right)[{(1-\frac{1}{a})}^n-(-1{)}^n.\frac{1}{a}]$

$f\left(2007\right)=a^{2006}\left[{\left(\frac{a-1}{a}\right)}^{2007}\right]-{(-1)}^{2007}.\frac{1}{a}$

$=a^{2006}.\left[\frac{3^{2007/223}}{a^{2007}}\right]+\frac{1}{a}$

$f\left(2008\right)=a^{2007}.\left[\frac{3^{2008/223}}{a^{2008}}\right]+\frac{1}{a}$

$f\left(2007\right)+f\left(2008\right)=\frac{1}{a}[3^{2007/223}+3^{2008/223}]$

$=[3^9+3^{9+1/223}]$

$=\frac{3^9(1+3^{1/223})}{a}$

$=\frac{3^9.a}{a}$

$\Rightarrow$ $3^9=9\times 2187$

$\Rightarrow$ $k=2187$.