chứng minh: 1/6<1/5^2+1/6^2+1/7^2+...+1/100^2<1/4
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Đặt \(B=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}\)
Ta thấy:
\(B=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
\(\Rightarrow B< \dfrac{1}{4}\)
Ta lại thấy:
\(B>\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}=\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{6}\)
\(\Rightarrow B>6\)
\(\Rightarrow\dfrac{1}{6}< B< \dfrac{1}{4}\left(dpcm\right)\)
a) ta có :1/5^2<1/4.5=1/4-1/5
1/6^2<1/5.6=1/5-1/6
.................
1/100^2<1/99.100=1/99-1/100
=>1/5^2+1/6^2+1/7^2+......+1/100^2 <1/4-1/100=6/25<1/4(1)
ta lại có:1/5^2>1/5.6=1/5-1/6
1/6^2>1/6.7=1/6-1/7
.................
1/100^2>1/100.101=1/100-1/101
=>1/5^2+1/6^2+1/7^2+......+1/100^2>1/5-1/101=96/505>1/6(2)
từ (1)(2) suy ra 1/6<1/5^2+1/6^2+1/7^2+......+1/100^2 < 1/4
b)ta có:1/11+1/12+....+1/70=(1/11+1/12+...+1/20)+(1/21+1/22+...+1/30)+(1/31+1/32+...+1/40)+(1/41+1/42+...+1/50)+(1/51+1/52+...+1/60)+(1/61+1/62+...+1/70)>(1/20+1/20+...+1/20)(10 phân số 1/20)+(1/30+1/30+...+1/30)(10 phân số 1/30)+(1/40+1/40+...+1/40)(10 phân số 1/40)+(1/50+1/50+...+1/50)(10 phân số 1/50)+(1/60+1/60+...+1/60)(10 phân số 1/60)=1/2+1/3+1/4+1/5+1/6=29/20>4/3(1)
ta lại có:1/11+1/12+....+1/70=(1/11+1/12+...+1/20)+(1/21+1/22+...+1/30)+(1/31+1/32+...+1/40)+(1/41+1/42+...+1/50)+(1/51+1/52+...+1/60)+(1/61+1/62+...+1/70)<(1/11+1/11+...+1/11)(10 phân số 1/11)+(1/21+1/21+...+1/21)(10 phân số 1/21)+(1/31+1/31+...+1/31)(10 phân số 1/31)+(1/41+1/41+...+1/41)(10 phân số 1/41)+(1/51+1/51+...+1/51)(10 phân số 1/51)+(1/61+1/61+...+1/61)(10phân số 1/61) =10/11+10/21+10/31+10/41+10/51+10/61=2,311777327<5/2(2)
từ (1)(2)=>4/3<1/11+1/12+....+1/70<5/2
1/5^2< 1/4.5=1/4-1/5
1/6^2<1/5.6=1/5-1/6
..
1/99^2<1/98.99=1/98-1/99
1/100^2<1/99.100=1/99-1/100
Cộng vế theo vế, đơn giản:
=> 1/5^2+1/6^2+...+1/100^2< 1/4 -1/100<1/4
**
1/5^2> 1/5.6=1/5-1/6
1/6^2>1/6.7=1/6-1/7
1/99^2>1/99.100=1/99-1/100
1/100^2>1/100.101=1/100-1/101
Cộng vế theo vế, đơn giản:
=> 1/5^2+1/6^2+...+1/100^2>1/5 -1/101=96/505>1/6
Vậy:
1/6<1/5^2+1/6^2+...+1/100^2<1/4.
Ta có: \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)(1)
\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{5}-\frac{1}{101}>\frac{1}{5}-\frac{1}{30}=\frac{1}{6}\)(2)
Từ (1) và (2) suy ra \(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)(đpcm)
Ta có:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)
Ta có:\(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{75};\frac{1}{76}>\frac{1}{77}>...>\frac{1}{100}\)
Tự giải tiếp hay nhờ thầy cô giảng tiếp đi nha bn, mỏi tay nên ko thể làm đc nữa !!
Số số x la : (52-2):2+1=26