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\(M=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{62}+\frac{1}{63}\)
\(M=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+\left(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}\right)+\left(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}\right)\)
\(M< 1+\frac{1}{2}.2+\frac{1}{4}.4+\frac{1}{8}.8+\frac{1}{16}.16+\frac{1}{32}.32\)
\(M< 1+1+1+1+1+1\)
\(M< 1.6=6\left(đpcm\right)\)
đpcm là điều phải chứng minh đúng không bn soyeon_Tiểubàng giải?
B < 1+1+1/2.3+1/3.4+...+1/62.63
B < 2+(1/2-1/3+1/3-1/4+...+1/62-1/63)
B < 2+(1/2-1/63)
B < 2+61/126 suy ra B < 2 và 6/126
Mà 2 + 61/126 <6
Suy ra B< 2+6/126<6 suy tiếp B < 6
Ta có:
\(\frac{1}{2}< 6\)
\(\frac{1}{3}< 6\)
\(...\)
\(\frac{1}{63}< 6\)
\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{63}< 6\)
\(\Rightarrow A< 6\left(dpcm\right)\)
\(#Jen\)
Trao đổi nếu cần
Trả lời
a) Đặt \(H=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(\Rightarrow H< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Leftrightarrow H< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow H< 1-\frac{1}{100}\)
\(\Leftrightarrow H< \frac{99}{100}\)
\(\Leftrightarrow A< 1+\frac{99}{100}\)
Ta thấy \(\frac{99}{100}< 1\Rightarrow A< 2\)
Vậy A<2 (đpcm)
b) Ta có: 1=1
\(\frac{1}{2}+\frac{1}{3}< \frac{1}{2}+\frac{1}{2}=1\)
\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}< \frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1\)
\(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+...+\frac{1}{15}< \frac{1}{8}+\frac{1}{8}+\frac{1}{8}+...+\frac{1}{8}=1\)
\(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}< \frac{1}{16}+\frac{1}{16}+...+\frac{1}{16}=1\)
\(\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}< \frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}=1\)
\(\Rightarrow B< 1+1+1+1+1+1\)
\(\Rightarrow B< 6\)
Vậy B<6 (đpcm)
a)A<1+1/1.2 +1/2.3 +1/3.4+...+1/99.100
A<1+1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A<2-1/100<2
b)B=1+1/2+(1/3+1/4)+(1/5+1/6+1/7+1/8)+(1/9+...+1/16)+(1/17+1/18+...+1/32)+(1/33+1/34+...+1/63+1/64)-1/64
B<1+1/2+1/2+1/2+1/2+1/2+1/2-1/64
B<1+3-1/64
B<4-1/64<6
a, \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(\Rightarrow1< 1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
Mà \(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1+1-\frac{1}{50}=2-\frac{1}{50}< 2\)
\(\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 2\Rightarrow A< 2\left(đpcm\right)\)
b, B = 2 + 22 + 23 +...+ 230
= (2+22+23+24+25+26)+...+(225+226+227+228+229+230)
= 2(1+2+22+23+24+25)+...+225(1+2+22+23+24+25)
= 2.63+...+225.63
= 63(2+...+225)
Vì 63 chia hết cho 21 nên 63(2+...+225) chia hết cho 21
Vậy B chia hết cho 21
