Cho M= 1/1! +1/2! +1/3! +......+ 1/100!
Chưng minh rằng : 3! - M > 4
Những câu hỏi liên quan
Chưng minh rằng:1/3^2+1/4^2+1/5^2+/6^2+....+1/100^2<1/2
Ta có: 1/3^2=1/3.3<1/2.3
1/4^2=1/4.4<1/3.4
1/5^2=1/5.5<1/4.5
1/6^2=1/6.6<1/5.6
...............................
1/100^2=1/100.100<1/99.100
=>1/3^2+1/4^2+1/5^2+1/6^2+....+1/100^2<1/2.3+1/3.4+1/4.5+1/5.6+....+1/99.100
=>1/3^2+1/4^2+1/5^2+1/6^2+....+1/100^2<1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+....+1/99-1/100
=>1/3^2+1/4^2+1/5^2+1/6^2+....+1/100^2<1/2-1/100
=>1/3^2+1/4^2+1/5^2+1/6^2+....+1/100^2<49/100 (1)
Ta có: 1/2=50/100>49/100 (2)
Từ (1) và (2) =>1/3^2+1/4^2+1/5^2+1/6^2+....+1/100^2<1/2(đpcm)
Đúng 0
Bình luận (0)
1.Chưng minh rằng (1+/1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh 1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
1.Chưng minh rằng
(1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
Xét: (1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100) =
(1+1/3+1/5+....+1/99) + (1/2+1/4+1/6+...+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1+1/2+1/3+...+1/50) =
1/51+1/52+1/53+ … + 1/100
Hay:
(1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
Viết lại:
(1+1/3+1/5+ … +1/199) – (1/2+1/4+1/6+ … +1/200) = 1/101+1/102+ … +1/200
Tương tự như trên ta được:
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1/2+1/4+1/6+...+1/200) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1+1/2+1/3+...+1/100) =
1/101+1/102+ … +1/200
Hay:
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
Đúng 0
Bình luận (0)
1 .Chưng minh rằng
(1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
Xét: (1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100) =
(1+1/3+1/5+....+1/99) + (1/2+1/4+1/6+...+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1+1/2+1/3+...+1/50) =
1/51+1/52+1/53+ … + 1/100
Hay:
(1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
Viết lại:
(1+1/3+1/5+ … +1/199) – (1/2+1/4+1/6+ … +1/200) = 1/101+1/102+ … +1/200
Tương tự như trên ta được:
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1/2+1/4+1/6+...+1/200) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1+1/2+1/3+...+1/100) =
1/101+1/102+ … +1/200
Hay:
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
Đúng 0
Bình luận (0)
1.Chưng minh rằng (1+/1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh 1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
1.Chưng minh rằng (1+/1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh 1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
1.Chưng minh rằng (1+/1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh 1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
1.Chưng minh rằng (1+/1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh 1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
1.Chưng minh rằng (1+/1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh 1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
1.Chưng minh rằng (1+/1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh 1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
1.Chưng minh rằng (1+/1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh 1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
1.Chưng minh rằng (1+/1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh 1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
ai tích mình tích lại
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
1.Chưng minh rằng (1+/1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh 1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
Chưng minh rằng :
\(\frac{1.2-1}{2\text{!}}+\frac{2.3-1}{3\text{!}}+\frac{3.\text{4}-1}{\text{4}\text{!}}+...+\frac{99.100-1}{100\text{!}}< 2\)
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=1-\frac{1}{2!}+1-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4!}+...+\frac{1}{98!}-\frac{1}{100!}\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
Vậy \(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\left(đpcm\right)\)
cho biểu thức M=2^3-1/2^3+1x3^3-1/3^3+1x4^3-1/4^3+1x...x100^3-1/100^3+1.Chứng minh rằng M>2/3
cho A=1/1!+1/2!+.....+1/100!. Chứng minh rằng 3!- M > 4
cho
M=\(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\)
Chứng minh rằng 3!-m>4
Cho M =\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\) .Hãy chứng minh M<\(\frac{3}{16}\)
Câu 2 Chứng minh rằng :
\(\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}< \frac{1}{50}\)
Tham khảo nha bạn :
Câu hỏi của Trần Minh Hưng - Toán lớp | Học trực tuyến
Đúng 0
Bình luận (0)
Chưng minh rằng :\(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{100^2}<1\)