cho A=1/2+1/3+..+1/100.chứng minh rằng A<3/4
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Chứng Minh Rằng
a. cho biểu thức A= 3 + 3^2+ 3^3+ 3^4+...+ 3^100 và B= 3^100-1.Chứng Minh rằng : A<B
b. Cho A= 1+4+4^2+...+4^99, B= 4^100. Chứng Minh Rằng : A<B/3
\(A=3+3^2+3^3+...+3^{100}\)
\(\Leftrightarrow3A=3^2+3^3+3^4+3^5+....+3^{101}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+3^5+...+3^{101}\right)-\left(3+3^2+3^3+3^4+...+3^{100}\right)\)
\(\Leftrightarrow2A=3^{101}-3\)
\(\Leftrightarrow A=\frac{3^{101}-3}{2}< 3^{100}-1\)
\(\Leftrightarrow A< B\)
a. tính A = 3+3^2+3^3+3^4+.....+3^100
3A=3^2+3^3+3^4+3^5+....+3^100
3A-A=(3^2+3^3+3^4+....+3^101)-(3+3^2+3^3+3^4+.....+3^100)=3^101-3=3^100
mà B=3^100-1 => A<B
\(A=1+4+4^2+...+4^{99}\)
\(\Leftrightarrow4A=4+4^2+4^3+...+4^{100}\)
\(\Leftrightarrow3A=4^{100}-1\)
\(\Leftrightarrow A=\frac{4^{100}-1}{3}< \frac{4^{100}}{3}\)
hay A<B (đpcm)
Bài 1:
a) Cho A = 1/2 + (1/2)^2 + (1/2)^3 +...+ (1/2)^99
Chứng minh rằng: A<1
b) Cho B = 1/3 + 2/3^2 + 3/3^3 + ... + 100/3^100
Chứng minh rằng: B<3/4
\(a.A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(2A-A=1-\frac{1}{2^{99}}\)
\(A=1-\frac{1}{2^{99}}< 1\)
\(b.B=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6A-2A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{203}{3^{100}}< 3\)
\(A< \frac{3}{4}\)
Ủng hộ mk nha ^_^
Cho A=1+\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2^{100}-1}\)
Chứng minh rằng 50<A<100
Cho A=1+1/2+1/3+1/4+....+1/(2100-1). Chứng minh rằng A < 100
Cho A= 1/2+1/2^2+1/2^3+1/2^4+1/2^100. Chứng minh rằng A<1
Cho B=2/1.2+2/2.3+2/3.4+...+2/99.100. chứng minh rằng c<2
A= \(\frac{1}{2}\) + \(\frac{1}{2^2}\) + \(\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\)
\(\Rightarrow\) 2A = 1 + \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\)
\(\Rightarrow\) 2A - A = ( \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\) ) -
( \(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\))
\(\Rightarrow\) A = 1 - \(\frac{1}{2^{100}}\) < 1
Vậy: A < 1
\(\frac{1}{2}\)
B= \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
= 2. \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
= 2. ( \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\) )
= 2. \(\left(\frac{1}{1}-\frac{1}{100}\right)\) = \(\frac{99}{50}\)
\(\Rightarrow\) B = \(\frac{99}{50}\) < \(\frac{100}{50}\) = 2
Vậy: B < 2
A=1/2^2+1/100^2 Chứng minh rằng A<1
B=1/1^2+1/1^2+1/3^2+...+1/100^2 Chứng minh rằng B<1 3/4 (hỗn số nhé)
C=1/1^2+1/4^2+1/6^2+...+1/100^2 Chứng minh rằng C<1/2
D=1/4^2+1/5^2+1/6^2+...+1/99^2+1/100^2 Chứng minh rằng 1/5<D<1/3
Giup mình nha mình đang cần gấp
a>
\(\frac{1}{2^2}+\frac{1}{100^2}\)=1/4+1/10000
ta có 1/4<1/2(vì 2 đề bài muốn chứng minh tổng đó nhỏ 1 thì chúng ta phải xét xem có bao nhiêu lũy thừa hoặc sht thì ta sẽ lấy 1 : cho số số hạng )
1/100^2<1/2
=>A<1
Cho a = 1+1/2+1/3+1/4+.....+1/(100-1). Chứng minh rằng 50<a<100
Cho: A= 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{2^{100}-1}\)
Chứng minh rằng: 50 < A < 100
Giúp mình với!
Cho a = 1+1/2+1/3+1/4+.....+1/(100-1)
Chứng minh rằng 50<a<100