1 + 3 + 9 + 27 + 6561 + 19683
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ta có :
= ( 1 + 59049 ) + ( 3 + 2187 ) + ( 9 + 6561 ) + ( 27 + 243 ) + ( 81 + 729 )
= 59050 + 2190 + 6570 + 270 + 810
= 59050 + ( 2190 + 810 ) + 6570 + 270
= 59050 + 3000 + 6570 + 270
= 59050 + ( 3000 + 6570 ) + 270
= 59050 + 9570 + 270
= 68620 + 270
= 68890
Bài làm
1 + 3 + 9 + 27 + 6561 + 19683
= ( 1 + 9 ) + ( 3 + 27 ) + ( 6561 + 19683 )
= 10 + 30 + 26244
= 40 + 26244
= 26284
cái này là toán lớp 6 mak!câu c có số nguyên âm nè!!!!!!!!!!
đặt A = 1+3+9+27+....+2187+6561
=>A = 30 + 31 + 32 + 33 + .. . +37 + 38
3A = 31 + 32 + 33 + ... + 38 + 39
3A - A = (31 + 32 + 33 + ... + 38 + 39)-(30 + 31 + 32 + 33 + .. . +37 + 38 )
2A = 39 - 1
A=\(\frac{3^9-1}{2}=\frac{19682}{2}=9841\)
\(3A=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{2187}\)
\(3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{2187}\right)-\left(\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{6561}\right)\)
\(2A=\dfrac{6560}{6561}\)
\(A=\dfrac{3280}{6561}\)
M = \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)
=> 3M = \(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{2187}\)
=> 3M - M = ( \(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{2187}\) ) - ( \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\))
2M = 1 - \(\frac{1}{6561}\)
2M = \(\frac{6560}{6561}\)
=> M = \(\frac{3280}{6561}\)
\(M=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+.......+\frac{1}{6561}\)
\(\Rightarrow M=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.........+\frac{1}{3^8}\)
\(\Rightarrow3M=3\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.........+\frac{1}{3^8}\right)\)
\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+............+\frac{1}{3^7}\)
\(\Rightarrow3M-M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..........+\frac{1}{3^7}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-.......-\frac{1}{3^8}\)
\(\Rightarrow2M=1-\frac{1}{3^8}\)
\(\Rightarrow M=\frac{1-\frac{1}{3^8}}{2}\)
Vậy M = \(\frac{1-\frac{1}{3^8}}{2}\)
\(A=\dfrac{3}{5.6}+\dfrac{3}{6.7}+...+\dfrac{3}{91.92}\)
\(\Rightarrow A=3\left(\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{91.92}\right)\)
\(\Rightarrow A=3\left(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{91}-\dfrac{1}{92}\right)\)
\(\Rightarrow A=3\left(\dfrac{1}{5}-\dfrac{1}{92}\right)\)
\(\Rightarrow A=3.\dfrac{87}{460}=\dfrac{261}{460}\)
1+3+9+27+6561+19683 = 26284
1 + 3 + 9 + 27 + 6561 + 19683
= ( 1 + 9 ) + ( 3 + 27 ) + ( 6561 + 19683)
= 10 + 30 + 26244
= 40 + 26244
= 26284
học tốt!!!