Chứng tỏ rằng 1/6 mũ 2 1/9 mũ 2 1/12 mũ ..... 1/30 mũ 2 <1/9
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\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{10-9}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}< 1\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\\ A< \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\\ A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}\\ A< \frac{9}{10}< 1\Rightarrow A< 1\)
Ta có 1/22<1/1.2
1/32<1/2.3
1/42<1/3.4
................
1/8²<1/7.8
=>B<1/1.2+1/2.3+1/3.4+...+1/7.8
=>B<1-1/2+1/2-1/3+1/3-1/4+...+1/7-1/8
=>B<1-1/8
Vậy B < 1
b=1/22+1/32+1/42+...+1/82<1/1.2+1/2.3+1/3.4+......+1/7.8
b=1-1/2+1/2-1/3+1/3-1/4+....+1/7-1/8
b=1-1/8
b=7/8
<=>b<1
k cho mink nha
b=1/22+1/32+1/42+...+1/82<1/1.2+1/2.3+1/3.4+......+1/7.8
b=1-1/2+1/2-1/3+1/3-1/4+....+1/7-1/8
b=1-1/8
b=7/8
<=>b<1
owo
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
\(\Rightarrow B=\frac{1}{2^2}+...+\frac{1}{8^2}< \frac{1}{1.2}+...+\frac{1}{7.8}\)
\(\Rightarrow\frac{1}{2^2}+...+\frac{1}{8^2}< 1-\frac{1}{2}+...+\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow\frac{1}{2^2}+...+\frac{1}{8^2}< 1-\frac{1}{8}\)
\(\Rightarrow\frac{1}{2^2}+...+\frac{1}{8^2}< \frac{7}{8}< 1\)
\(\Rightarrow B< 1\)
P = 32 + 62 + 92 + ... + 302
P = 32 . (12 + 22 + 32 + ... + 102)
P = 9 . 385
P = 3465
a) C = 106 + 57
C = 26 . 56 + 57
C = 56 . (26 + 5)
C = 56 . (64 + 5)
C = 56 . 69 chia hết cho 69
b) 310 . 199 - 39 . 500
= 39 . (3.199 - 500)
= 39 . (597 - 500)
= 39 . 97 chia hết cho 97
Có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{8^2}< \frac{1}{7.8}\)
\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}\)
\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow B< 1-\frac{1}{8}< 1\)
\(\Rightarrow B< 1\) \(\Rightarrowđpcm\)