Tính tổng dãy số S= 2×4 +4×6+ … +1998×2000?

0 bình luận về “Tính tổng dãy số S= 2×4 +4×6+ … +1998×2000?”

1. Ta có

   $S = 2\times 4 + 4 \times 6 + \cdots + 1998 \times 2000$

   $= 2 \times 1 \times 2 \times 2 + 2 \times 2 \times 3 \times 2 + \cdots + 2 \times 999 \times 1000 \times 2$

   $= 4(1 \times 2 + 2 \times 3 + \cdots + 999 \times 1000)$

   Đặt

   $A = 1 \times 2 + 2 \times 3 + \cdots + 999 \times 1000$

   Nhân $3$ vào 2 vế ta có

   $3A = 1 \times 2 \times 3 + 2 \times 3 \times 3 + \cdots + 999 \times 1000 \times 3$

   $= 1 \times 2 \times 3 + 2 \times 3 \times (4-1) + 3 \times 4 \times (5-2) + \cdots + 999 \times 1000 \times (1001 – 998)$

   $= (1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + \cdots + 999 \times 1000 \times 1001) – (1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + \cdots + 998 \times 999 \times 1000)$

   $= 999 \times 1000 \times 1001$

   Suy ra

   $A = 333 \times 1000 \times 1001$

   Vậy ta có

   $S = 4A = 333 \times 4000 \times 1001$

   Bình luận

2. Ta có:

   S = 2×4+4×6+⋯+1998×2000S=2×4+4×6+⋯+1998×2000

    = 2 × 1 × 2 × 2 + 2 × 2 × 3 × 2 +⋯+ 2 × 999 × 1000 × 2 = 2 × 1 × 2 × 2 + 2 × 2 × 3 × 2 +⋯+ 2 × 999 × 1000 × 2

    = 4 (1 × 2 + 2 × 3 +⋯+ 999 × 1000) = 4(1 × 2 + 2 × 3 +⋯+ 999 × 1000)

   A = 1 × 2 + 2 × 3 +⋯+ 999 × 1000

   Nhân 3 vào 2 vế ta có:

   A = 1 × 2 × 3 + 2 × 3 × 3 +⋯+ 999 × 1000 × 3

   = 1 × 2 × 3 + 2 × 3 × (4 − 1) + 3 × 4 × (5 − 2) +

   ⋯+ 999 × 1000 × (1001 − 998) = 1 × 2 × 3 + 2 × 3 × (4 − 1) + 3 × 4 × (5 – 2) +⋯+ 999 × 1000 × (1001 − 998)

   = (1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 +⋯+ 999 × 1000 × 1001) − (1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 +⋯+ 998 × 999 × 1000) = (1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 +⋯+ 999 × 1000 × 1001) − (1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 +⋯+ 998 × 999 × 1000)

   = 999 × 1000 × 1001 = 999 × 1000 × 1001

   Suy ra:

   A = 333 × 1000 × 1001

   Vậy ta có:

   S = 4

   -////-

   Bình luận