Sum of Complex Numbers

In general, for $f(2N,N)$ , where $N$ is a non-zero real number, we have:

$f(2N,N)=\frac{2N+N \cdot i}{2N-N \cdot i}=\frac{2+i}{2-i}=\frac{(2+i)^2}{5}$

Analagously, we have:

$f(2,1)+f(4,2)+f(6,3)+f(8,4)+f(10,5)=5*\frac{(2+i)^2}{5}=(2+i)^2$

$=3+4i=C+Di$

Then our desired value is $C+D=3+4=\boxed{7}$

f(A,B)=(A+Bi)²/(A²+B²)=(A²-B²+2ABi)/(A²+B²). Making A=2K,B=K we get f(2k,k)=3k²/5k²+4k²i/5k²=3/5+4i/5. Then C+Di=5f(2k,k)=3+4i---> Answer=3+4=7

you may notice that each and every term of this sequence is equal to ( (2+i)/(2-i)) which on further simplification gives (( 3 +4i )/5)............ hence the sum of five such terms is(5*((3+4i)/5)) which is equal to (3+ 4i)