C = 1/3^1+2/3^2+...+100/3^100
Chứng minh C < 3/4
giúp mk nha các bn
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
C = 1/3^1 + 2/3^2 + .3/3^3 + .. + 100/3^100
1/3 . C = 1/3^2 + 2/3^3 + 3/3^4 + .. + 100/3^101
=> C - 1/3 . C = 1/3^1 + (2/3^2 - 1/3^2) + (3/3^3 - 2/3^3) + ... +(100/3^100 - 99/3^100) - 100/3^101
=> 2/3. C = 1/3^1 + 1/3^2 + 1/3^3 + .. + 1/3^100 - 100/3^101
xét S= 1/3^1 + 1/3^2 + 1/3^3 + .. + 1/3^100 tương tự
1/3 . S = 1/3^2 + 1/3^3 + 1/3^4 + .. + 1/3^101
=> S - 1/3. S = 1/3^1 - 1/3^101
=> 2/3. S = (1/3 - 1/3^101)
=> S = 3/2. (1/3 - 1/3^101) thay vào C ta có
2/3. C = 3/2. (1/3 - 1/3^101) - 100/3^101
=> C = 9/4. (1/3 - 1/3^101) - 150/3^101
=> C = 3/4 - 9/4*1/3^101 - 150/3^101 < 3/4
Ta co :
1/2! +2/3! +3/4! +... + 99/100!
= (1/1! -1/2!) + (1/2! - 1/3!) + (1/3! -1/4!) + .... + (1/99! -1/100!)
=1 - 1/100! <1
lik e nhe
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)