Question 1050433
{{{i^(4k) = (i^4)^k = 1^k = 1}}}
===>{{{i^(4k+1) = i}}},
{{{i^(4k+2) = -1}}}, and 
{{{i^(4k+3) = -i}}}.

===> {{{(4k+1)i^(4k) + (4k+2)i^(4k+1) + (4k+3)i^(4k+2) + (4k+4)i^(4k+3)}}}

= {{{(4k+1)*1 + (4k+2)*i + (4k+3)*(-1) + (4k+4)*(-i)}}}

= {{{-2 - 2i}}}.


Now substitute the values of k = 0, 1, 2, 3,...,497, 498 into the formula.

On the left side of the equation, we would get

{{{1+2i+3i^2+4i^3}}}+...+{{{1995i^1994+1996i^1995}}}, 

while on the right side, we would get

499*(-2 - 2i) = -998 - 998i.

Therefore,

{{{1+2i+3i^2+4i^3}}}+...+{{{1995i^1994+1996i^1995 = -998-998i}}}.