Evaluate `sum_(n=1)^(13)(i^n+i^(n+1)),` where `n in Ndot`

Explanation

Given that, `sum_(n=1)^(13) (i^(n) + i^(n+1)), "where n" in N. `
` =(i+i^(2)+ i ^(3) + i^(4)+ i^(5)+ i^(6)+ i^(7)+ i^(8)+ i^(9)+ i^(10)+ i^(11)+ i^(12)+ i^(13))` + ` (i+i^(2)+ i ^(3) + i^(4)+ i^(5)+ i^(6)+ i^(7)+ i^(8)+ i^(9)+ i^(10)+ i^(11)+ i^(12)+ i^(13)+i^(14))`
` =(i+2i^(2)+ 2i ^(3) + 2i^(4)+ 2i^(5)+ 2i^(6)+ 2i^(7)+ 2i^(8)+ 2i^(9)+ 2i^(10)+ 2i^(11)+ 2i^(12)+ 2i^(13)+i^(14))`
=`i-2-2i+2+2i+2(i^(4))+i^(2)+2(i)^(4)i^(3)+2(i^2)^(4)i+2(i^(2))^(5)+2(i^(2))^(5).i+2(i^(2))^(6)+2(i^(2))^(6).i+(i^(2))^(7)`
`=i-2-2i+2+2i-2-2i+2+2i-2-2i+2+2i-1-1-i`
"Alternate Method"
`sum_(n=1)^(13) (i^(n) + i^(n+1)), " n" in N =`
`=(1 -i)[i+i^(2)+i^(3)+i^(4)+i^(5)+i^(6)+i^(7)+i^(8)+i^(9)+i^(10)+i^(11)+i^(13)] `
`=0(1-i)[i^(13)]" "[:.i^(n)+i^(n+1)+i^(n+2)+i^(n+3)= 0. where n in N i.e., `sum_(n=1)^(13) i^(n) =0`
=`(+i)i" "[:.(i^(4)^(3).i = i] `
=`(i^(2)+i)=i-1`