cho s =3/1*4+3/4*7+3/7*10+......+3/43*46.chung minh rang s <1
Moi nguoi tra loi thu cau hoi nay nhe ( de thoi ma )
Cho S= 3/1*4+3/4*7+3/7*10+....+3/40*43+3/43*46
Hay chung to rang S < 1
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}=\frac{45}{46}\)
S=1-1/4+1/4-1/7+1/7-1/10+...+1/43-1/46 S=1-1/46 S=45/46 ok S<1
Cho S = 3/1*4+3/4*7+3/7*10+...+3/40*43+3/43+46. Hãy chứng tỏ S < 1
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\) < 1
\(S=3\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{40.43}+\frac{1}{43.46}\right)\)
\(S=3.\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\right)\)
\(\Rightarrow S=1-\frac{1}{46}\Rightarrow S< 1\left(đpcm\right)\)
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
= \(1-\frac{1}{46}< 1\)
\(\Rightarrow S< 1\left(đpcm\right)\)
3/4*1+3/4*7+3/7*10+........+3/40*43+3/43*46 chứng mính s<1
Ta có: \(\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{40.43}+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}\)
Vì \(\frac{1}{46}>0\Rightarrow1-\frac{1}{46}< 1\)
Vậy \(\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{43.46}< 1\)
ChoS= 3/1*4 +3/4*7 +3/7*10 +.......+ 3/40*43 +3/43*46
Chứng tỏ S <1
\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}< 1\)
Vậy \(S< 1\)
Chúc bạn học tốt !!!
cho s=3/1*4+3/4*7+...3/40*43+3/43*46
S=3.(1/1-1/4+1/4-1/7+.........+1/40-1/43+1/43-1/46)
S=3.(1/1-1/46)
S=3.45/46
S=2/43/46
S = 3/1*4 + 3/4*7 + 3/7*10 + ...+3/40*43 + 3/43*46
hãy chứng tỏ S < 1
Các bạn giúp dùm mình với nhé !
giúp mình với đi
các bạn ơi
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}< 1\)
Chứng tỏ S < 1
Ủng hộ mk nha ^_^
S = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}=\frac{45}{46}< 1\)
S=1-1/4 + 1/4 - 1/7 +1/7 - 1/10 +...+ 1/40 - 1/43 +1/43 -1/46
= 1-1/46 < 1 (ĐPCM)
Cho S = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7\cdot10}+....+\frac{3}{40\cdot43}+\frac{3}{43\cdot46}\) . Hay chung to rang S < 1
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}\)
\(s=1-\frac{1}{46}< 1\)
Vậy S<1
\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{43\cdot46}\)
\(S=1\left[\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{43\cdot46}\right]\)
\(S=1\left[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}\right]\)
\(S=1\left[1-\frac{1}{46}\right]=1\cdot\frac{45}{46}=\frac{45}{46}< 1(đpcm)\)
S = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
S = \(\frac{1}{1}-\frac{1}{46}\)
S = \(\frac{46}{46}-\frac{1}{46}\)
S = \(\frac{45}{46}< 1\)
Hay S < 1
Vậy S < 1 (đpcm)
Cho S = 3/1x4 + 3/4x7 + 3/7×10 +......+ 3/40×30 + 3/43×46. Hãy chứng tỏ S<1
Ai nhanh mk tích
S=3/1.4+3/4.7+3/7.10+.....+3/40.43+3/43.46
S= 1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46
S= 1-1/46
=> S<1
S=3.(1/1-1/4+1/4-1/7+.........+1/40-1/43+1/43-1/46)
S=3.(1/1-1/46)
S=3.45/46
S=2/43/46
=> 2/43/46>1
=>S>1
chung minh rang 4+4^3+4^5+4^7+...+4^23 chia het cho 68
chung minh rang 1+3+3^2+3^3+...+3^2000 chia het cho 13
giup mink voi thu 6 mink nop roi
4 + 4^3 + 4^5 + 4^7 + ... + 4^23
= ( 4 + 4^3 ) + ( 4^5 + 4^7 ) +.....+ ( 4^22 + 4^23)
=4( 1+16 ) + 4^5( 1+16 ) +....+ 4^22( 1+ 16 )
=4 x 17 + 4^5 x 17+....+ 4^22 x 17 chia hết cho 68
Câu 2:
1+3+3^2+3^3+....+3^2000
=( 1+3 +3^2 ) + ( 3^3 + 3^4 + 3^5 ) +.....+ ( 3^ 1998 + 3^1999 + 3^2000)
=1( 1+ 3 + 9 ) + 3^3 + ( 1+ 3 + 9 ) +......+ 3^1998+( 1+ 3 + 9 )
= 1 x 13+ 3^3 x 13 +......+ 3^1998 x 13 chia hết cho 13
k mk nha lần sau mk k lại
Câu 1 nha : 4+4^3+4^5+4^7+....+4^23 = (4+4^3)+(4^5+4^7)+....+(4^21+4^23)
= 68 + 4^4.(4+4^3)+....+4^20.(4+4^3) = 68 + 4^4.68 + .... + 4^20.68
=68.(1+4^4+....+4^20) chia hết cho 68
Câu 2 nha 1+3+3^2+...+3^2000 = (1+3+3^2)+(3^3+3^4+3^5)+....+(3^1998+3^1999+3^2000)
= 13 + 3^3.(1+3+3^2)+....+3^1998.(1+3+3^2) = 13+3^3.13+....+3^1998.13
=13.(1+3^3+....+3^1998) chia hết cho 13