Let `a=3^(1/(223))+1` and for all `n geq3,l e tf(n)=^n C_0dota^(n-1)-^n C_1dota^(n-2)+^n C_2dota^(n-3)-+(-1)^(n-1).^n C_(n-1) a^0` . If the value of `f(2007)+f(2008)=3^k` where k in N , then the value of `k` is.

Answer

9

Explanation

`f(n)=.^(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)`
`= 1/a(.^(n)C_(0)a^(n)-.^(n)C_(1)a^(n-1)+.^(n)C_(2)a^(n-2)+"...."+(-1)^(n-1)C_(n-1)a)`
`=1/a((a-1)^(n)-(-1)^(n).^(n)C_(n))`
`= 1/a((3^(n/223)-(-1)^(n)))`
`f(x) = (3^(n/223)-(-1)^(n))/((3^(1/223)+1))`
`rArr f(2007)= (3^(2007/223)+1)/(3^(1/223)+1)`
`rArr f(2008)= (3^(2008/223)+1)/(3^(1/223)+1)`
`f(2007)+f(2008)= (3^(2007/223)+3^(2008/223))/(3^(1/223+1))`
`= (3^(9)+3^(9+(1)/(223)))/(3^(1/223)+1)`
`= 3^(9)((1+3^(1/223)))/((1+3^(1/223)))=3^(9)`
`rArr 3^(9)= 3^(k)` then `k = 9`