So sánh
1/1×2×3 + 1/2×3×4 + 1/3×4×5 +...... 1/ 23×24×25 và 1/4
1/4 + 1/16+ 1/36 + 1/64 + 100 +1/144 + 1/196 và 1/2
Tính:
A=1*2*3+2*3*4+3*4*5+......+98*99*100
CMR:
A=1/4+1/16+1/36+1/64+1/100+1/144+1/196<1/2
A=\(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}\)=\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
=>A<\(\frac{1}{2.2}+\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}+\frac{1}{12.14}\)
=>A<\(\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{12}-\frac{1}{14}\right)\)\(:2\)=\(\left(\frac{1}{2}-\frac{1}{14}\right):2\)<\(\frac{1}{2}\)
=>A<\(\frac{1}{2}\)
1 rút gọn 1/1 X 3 + 1/3 X 5 + 1/5 X 7+......+ 1/(2n+1) X (2n+3)
2 so sánh 1/2 X 3/4 X 5/6 X...... X 9999/10000 và 1/100
3 Tìm số tự nhiên x biết
1/3 +1/6 +1/10+....+2/x(x+1)=2017/2019
4 Chứng minh rằng
1/4 +1/16 +1/36 +1/64 +1/100 +1/144 +1/196 < 1/2
Chứng minh
a, 1/4+1/16+1/36+1/64+1/100+1/144+1/196<1/2
b, 11/15<1/21+1/22+1/23+...+1/59+1/60<3/2
a,\( \dfrac{1}{4}+ \dfrac{1}{16}+ \dfrac{1}{36}+ \dfrac{1}{64}+ \dfrac{1}{100}+ \dfrac{1}{144}+ \dfrac{1}{196}\)
= \( \dfrac{1}{4}+ \dfrac{1}{16}+ \dfrac{1}{36}+...+ \dfrac{1}{196} < \dfrac{1}{2^2-1}+ \dfrac{1}{4^2-1}+ \dfrac{1}{6^2-1}+...+ \dfrac{1}{14^2-1}\)
= \( \dfrac{1}{1.3}+ \dfrac{1}{3.5}+ \dfrac{1}{5.7}+...+ \dfrac{1}{13.15}\)
= \( \dfrac{1}{2}(1- \dfrac{1}{3}+ \dfrac{1}{3}- \dfrac{1}{5}+ \dfrac{1}{5}- \dfrac{1}{7}+ \dfrac{1}{7}-...- \dfrac{1}{13}+ \dfrac{1}{13}- \dfrac{1}{15})\)
= \( \dfrac{1}{2}(1- \dfrac{1}{15})< \dfrac{1}{2}\)
Vậy \( \dfrac{1}{4}+ \dfrac{1}{16}+ \dfrac{1}{36}+ \dfrac{1}{64}+ \dfrac{1}{100}+ \dfrac{1}{144}+ \dfrac{1}{196}\) \(<\dfrac{1}{2} \)
b,A= \(\dfrac{11}{15}<\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+...+\dfrac{1}{59}+\dfrac{1}{60}<\dfrac{3}{2}\)
\(=(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+....+\dfrac{1}{40})+(\dfrac{1}{41}+...+1...\)
\(=(\dfrac{20}{20.21}+\dfrac{21}{21.22}+...+\dfrac{39}{39.40})+(40/...\)
\(20(\dfrac{1}{20.21}+\dfrac{1}{21.22}+...\dfrac{1}{39.40})+40(\dfrac{1}{40}...\)
\(20(\dfrac{1}{20}-\dfrac{1}{40})+40(\dfrac{1}{40}-\dfrac{1}{60})>\dfrac{11}{15}\)
Lại có \(A<40(\dfrac{1}{20.21}+...\dfrac{1}{39.40})+60(\dfrac{1}{40.41}+...+...\)
\(=40(\dfrac{1}{20}-\dfrac{1}{40})+60(\dfrac{1}{40}-\dfrac{1}{60})<\dfrac{3}{2}\)
=> \(\dfrac{11}{15}<\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+...+\dfrac{1}{59}+\dfrac{1}{60}<\dfrac{3}{2}\)
So sánh \(\frac{1}{4} + \frac{1}{16} + \frac{1}{36} + \frac{1}{64} + \frac{1}{100} + \frac{1}{144} + \frac{1}{196}\) và \(\frac{1}{2}\)
\(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}\)
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{7^2}\right)\)
\(< \frac{1}{2^2}\left(1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{1}{2^2}\left(1-\frac{1}{7}\right)\)
\(=\frac{1}{2^2}\cdot\frac{6}{7}\)
\(=\frac{3}{14}\)
\(< \frac{1}{2}\)
so sanh 1/4+1/16+1/36+1/64+1/100+1/144+1/196 va 1/2
so sánh 2 lũy thừa 3^4 và 9^3
A=1+2+2^2+2^3+...+2^2017 và B=2^2018-1
16^19 và 8^25
5^23 và 6x5^22
5^36 và 11^24
Chứng minh:
c.\(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{59}+\frac{1}{60}< \frac{3}{2}\)
b.\(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+\frac{1}{41}+\frac{1}{61}+\frac{1}{85}+\frac{1}{113}< \frac{1}{2}\)
a.\(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}< \frac{1}{2}\)
1. chứng minh
\(\dfrac{1}{4}\)+ \(\dfrac{1}{16}\)+ \(\dfrac{1}{36}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{100}\)+\(\dfrac{1}{144}\)+\(\dfrac{1}{196}\)≤\(\dfrac{1}{2}\)
2.tìm số nguyên n sao cho (3n +24) \(⋮\) (n-4)
cứu t zới!!!huhu
Giải thích các bước giải:
Đặt A= 1/4+1/16+1/36+1/64+1/100+1/144+1/196
= 1/2^2+ 1/4^2+ 1/6^2+….+ 1/16^2
= 1/2^2.( 1+ 1/2^2+ 1/3^2+…+ 1/8^2)
Ta có 1+ 1/2^2+ 1/3^2+…+ 1/8^2< 1+ 1/1.2+ 1/2.3+….7.8= 1+ 1-1/2+ 1/2- 1/3+….+ 1/7- 1/8
= 2- 1/8< 2
Nên ( 1+ 1/2^2+ 1/3^2+…+ 1/8^2)< 2
=> A< 1/2^2 nhân 2= 1/2
Vậy A< 1/2
chứng minh rằng a 1/4 +1/16+1/36+1/64+1/100+1/144 +1/196+......+1/10000 <1/2