The powers of i repeat in blocks of four:
i,i2,i3,i4=1,−1,−i,i
Therefore, the sum telescopes:
\(\begin{align*} i + 2i^2 + 3i^3 + 4i^4 + \dots + 64i^{64} &= i + 2(-1) + 3(-i) + 4i + \dots + 64i \\ &= (i - 2i + 3i - 4i + \dots + 64i) + (2 - 3 + 4 - \dots + 64) \\ &= 0 + (2 + 64) \cdot 32 \\ &= \boxed{496}. \end{align*}\)
+758 As of what the first guest, i has four different stages. These stages are i, -1, -i, 1.Through these stages, we can see that i + 2i2 + 3i3 + 4i4 is equal to 2 - 2i. As we continue to 8i8, the answer is 4 - 4i. We can see a pattern, which is that for each four stages, we get 2 - 2i, no matter what. 64 / 4 is 16, so therefore the answer is 16 * (2 - 2i), or 32 - 32i.
{( 0 +1i ) , ( -2 ) , ( 0 -3i ) , ( 4 ) , ( 0 +5i ) , ( -6 ) , ( 0 -7i ) , ( 8 ) , ( 0 +9i ) , ( -10 ) , ( 0 -11i ) , ( 12 ) , ( 0 +13i ) , ( -14 ) , ( 0 -15i ) , ( 16 ) , ( 0 +17i ) , ( -18 ) , ( 0 -19i ) , ( 20 ) , ( 0 +21i ) , ( -22 ) , ( 0 -23i ) , ( 24 ) , ( 0 +25i ) , ( -26 ) , ( 0 -27i ) , ( 28 ) , ( 0 +29i ) , ( -30 ) , ( 0 -31i ) , ( 32 ) , ( 0 +33i ) , ( -34 ) , ( 0 -35i ) , ( 36 ) , ( 0 +37i ) , ( -38 ) , ( 0 -39i ) , ( 40 ) , ( 0 +41i ) , ( -42 ) , ( 0 -43i ) , ( 44 ) , ( 0 +45i ) , ( -46 ) , ( 0 -47i ) , ( 48 ) , ( 0 +49i ) , ( -50 ) , ( 0 -51i ) , ( 52 ) , ( 0 +53i ) , ( -54 ) , ( 0 -55i ) , ( 56 ) , ( 0 +57i ) , ( -58 ) , ( 0 -59i ) , ( 60 ) , ( 0 +61i ) , ( -62 ) , ( 0 -63i ) , ( 64 ) } = 32 - 32i
