`hat i xx(vec a xx hat i) + hat j xx (vec a xx hat j) + hat k xx (vec a xx hat k)` is equal to

Options

A. 0
B. 1
C. 2
D. More than 2

Explanation

`d_(n+1) =2d_(n)+n(1+2^(n))`
Dividing with `2^(n+1)` on both sides.
`(d_(n+1))/(2^(n+1))=(d_(n))/(2^(n))+(n)/(2)(2^(-n)+1)`
` therefore (d_(n+1))/(2^(n+1))-(d_(n))/(2^(n))=(n-1)/(2)(2^(-(n-1))+1)`
`(d_(2))/(2^(2))-(d_(1))/(2^(1))=(1)/(2)(2^(-1)+1)`
Adding, we get `d_(n+1)=(n^(2)+n+6).2^(n-1)-n-2`
`therefore d_(n)=((n-1)^(2)+(n-1)+6)2^(n-2)-(n-1)-2`
`d_(n)=(n^(2)-n+6)2^(n-2)-n-1`
Comparing both `d_(n)` 's we get
`n^(2)-2n+13=n^(2)-n+6`
`impliesn=7`