The number of `n` digit number formed by using digits `{1,2,3}` such that if `1` appears, it appears even number of times, is

Options

A. `2^(n)+1`
B. `(1)/(2)(3^(n)+1)`
C. `(1)/(2)(3^(n)-1)`
D. `(1)/(2)(2^(n)-1)`

Answer

B

Explanation

`(b)` Required number of numbers `="(n)C_(0)(2)^(n)+"(n)C_(2)(2)^(n-2)+…`
We know that
`(1+x)^(n)+(x-1)^(n)=2sum_(k=0)^(n)"(n)C_(k)x^(k)`
Hence, put `x=2`, get the result
`2["^(n)C_(0)(2)^(n)+^(n)C_(2)(2)^(n-2)+^(n)C_(2)(2)^(n-4)+^(n)C_(6)(2)^(n-6)+...]`