Chứng minh rằng: 1/22 + 1/32 + 1/42 + ........+1/20132 <1
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Những câu hỏi liên quan
Chứng minh rằng: M = 1/22 + 1/32 + 1/42 + ... + 1/n2 < 1
Chứng minh rằng:
A = 1/3 + 1/32 + 1/33 + ..........+ 1/399 < 1/2
B = 3/12x 22 + 5/22 x 32 + 7/32 x 42 +............+ 19/92 x 102 < 1
C = 1/3 + 2/32 + 3/33 + 4/34 +.........+ 100/3100 ≤ 0
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
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Tính
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Đọc tiếp
Tính C = 1 2 – 2 2 + 3 2 – 4 2 + 5 2 – 6 2 + … . + 2013 2 – 2014 2 + 2015 2
C=12-22+32-42+52-62+..+20132-20142+20152
SSH:(20152-12):10+1=2015
(12-22)+(32-42)+(52-62)+...+(20132-20142)+20152
-10+(-10)+(-10)+...+(-10)+20152
-10x(2015-1):2+20152=12
=> C=12
Câu 1: Chứng minh rằng 2 + 1/2 +1/3 + 1/4 +...+ 1/67 > 5
Câu 2: Cho B = 1/22 +1/32 +...+ 1/902
Chứng minh 42/91 < B < 1
Câu 1: Chứng minh rằng 2 + 1/2 + 1/3 + 1/4 +...+ 1/67 > 5
Câu 2: Cho B = 1/22 + 1/32 +... + 1/902
Chứng minh: 42/91 < B < 1
chứng mỉnh rằng 1/22 +1/32 +1/42 + ...+ 1/92 <8/9
Ta thấy:
\(2^2=2.2>1.2\Rightarrow\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(3^2=3.3>2.3\Rightarrow\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
.................
\(9^2=9.9>8.9\Rightarrow\dfrac{1}{9^2}< \dfrac{1}{8.9}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}=1-\dfrac{1}{9}=\dfrac{8}{9}\)
=> Đpcm
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Ta thấy:
22=2.2>1.2⇒122<11.222=2.2>1.2⇒122<11.2
32=3.3>2.3⇒132<12.332=3.3>2.3⇒132<12.3
.................
92=9.9>8.9⇒192<18.992=9.9>8.9⇒192<18.9
⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9
⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89
=> ...(tự viết)
Ta thấy:
22=2.2>1.2⇒122<11.222=2.2>1.2⇒122<11.2
32=3.3>2.3⇒132<12.332=3.3>2.3⇒132<12.3
.................
92=9.9>8.9⇒192<18.992=9.9>8.9⇒192<18.9
⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9
⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89
=> 11111111111111111111110101010110000
HACK
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cho sửa lại là:
Ta thấy:
\(⇒ \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... < \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(⇔\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2} <1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4} ...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
⇒ {{{tự làm nhé}}}
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hãy chứng minh: 1/22+1/32+1/42+...+1/20212+1/20222<1
chứng minh
1/22+1/32+1/42+1/52+...+1/1002 >3/4
Bài) Chứng minh rằng
50/51<1+1/22+1/32+1/42+...+1/502<2