Cho S= \(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}+\frac{1}{49}+\frac{1}{64}+\frac{1}{81}\)
Chứng minh rằng S < \(\frac{1}{2}\)
Giúp mình, mk cần gấp. Bạn nào nhanh mình tick cho
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\(\frac{1}{5}+\frac{4}{10}+...+\frac{81}{45}=\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+...+\frac{9}{5}=\frac{1+2+3+...+9}{5}=\frac{45}{5}=9\)
1/5 + 4/10 + 9/15 + 16/20 + 25/25 + 36/30 + 49/35 + 64/40 + 81/45
=1/5 + 2/5 + 3/5 + 4/5 + 5/5 + 6/5 + 7/5 + 8/5 + 9/5
=45/5 = 9
#)Giải :
\(\frac{1}{15}+\frac{4}{30}+\frac{9}{45}+\frac{16}{60}+...+\frac{81}{135}=\frac{1}{15}+\frac{2}{15}+\frac{3}{15}+...+\frac{9}{15}=\frac{45}{15}=3\)
Dễ ẹc ak :v rút gọn là ra
=(\(\frac{1}{15}\)+\(\frac{4}{30}\)+\(\frac{16}{60}\)+\(\frac{64}{120}\))+(\(\frac{9}{45}\)+\(\frac{36}{90}\))+(\(\frac{25}{75}\)+\(\frac{81}{135}\))
=(\(\frac{8}{120}\)+\(\frac{16}{120}\)+\(\frac{32}{120}\)+\(\frac{64}{120}\))+(\(\frac{18}{90}\)+\(\frac{36}{90}\))+\(\frac{14}{15}\).
=1+\(\frac{3}{5}\)+\(\frac{14}{15}\).
=\(\frac{8}{5}\)+\(\frac{14}{15}\).
=\(\frac{15}{38}\)
1/10+4/20+9/30+16/40+25/50+36/60+49/70+64/80+81/90
=1/10+2/10+3/10+4/10+5/10+6/10+7/10+8/10+9/10
=1+2+3+4+5+6+7+8+9/10
=45/10 (tự rút gọn)
a) 5/30+15/90+25/150+35/210+45/270
=1/6+1/6+1/6+1/6+1/6
=1/6 x 5
=5/6
b) 1/2+1/6+1/12+1/20+....+1/56
=1/1x2+1/2x3+1/3x4+1/4x5+.....1/7x8
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......-1/7+1/7-1/8
=1/1-1/8
=7/8
c) mình chịu
\(có\) \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}\approx1,4\)
\(mà\) \(\frac{1}{2}=1,5\)
\(=>\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}<\frac{1}{2}\)
\(\frac{1}{4}+\frac{1}{16}+...+\frac{1}{196}\)\(<\frac{1}{2^2-1}+\frac{1}{4^2-1}+\frac{1}{6^2-1}+...+\frac{1}{14^2-1}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{15}\right)<\frac{1}{2}\) \(\left(đpcm\right)\)