S=2/2.6+2/6.10+...+2/96.100. Tính tổng
S=2/2.6+2/6.10+...+2/96.100
Tính tổng:
S=\(\dfrac{2}{2.6}\)+\(\dfrac{2}{6.10}\)+\(\dfrac{2}{10.14}\)+...+\(\dfrac{2}{96.100}\)
\(S=\dfrac{2}{2\cdot6}+\dfrac{2}{6\cdot10}+...+\dfrac{2}{96\cdot100}\\ =\dfrac{1}{2}\left(\dfrac{4}{2\cdot6}+\dfrac{4}{6\cdot10}+...+\dfrac{4}{96\cdot100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{10}+...+\dfrac{1}{96}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=\dfrac{1}{2}\cdot\dfrac{49}{100}\\ =\dfrac{49}{100}\)
Tính tổng:
A=\(\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{10.14}+.....+\frac{1}{98.102}\)
B=\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{100^2}\)
Tính bằng máy casio nka các pạn..theo công thức tổng xích ma hay j đó nka
bai1:tinh tong S=1.3+3.5+5.7+...+99.101
bai2 :tinh tong S=1.4+4.7+7.10+...+2017.2020
bai 3: tinh tong N=2.4+4.6+6.8+..+100.102
bai 4: tinh tóng=2.6+6.10+10.14+14.18+...+42.46+50.54
bai 5:tinh tongB=2^2+4^2+6^2+...+100^2
bai 6:C=1^2+3^2+...+100^2
bai7: biet 1^2+2^2+3^2+...+10^2=385 tinh tong 2^2+4^2+6^2+...+20^2
bai 8: tinh tong s=1^2+2^2+3^2+...+99^2
Bài 1 :
\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)
Ta thấy :
\(3=2^2-1\)
\(15=4^2-1\)
\(35=6^2-1\)
.....
\(9999=100^2-1\)
\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)
\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)
\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)
nhanh len nhé mik đang cần gấp ai lam trước mik tích cho
Bài 6 :
\(C=1^2+2^2+...+100^2=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}=\dfrac{100.101.201}{6}=338350\)
Bài 9 :
\(S=1^2+2^2+3^2+...+99^2=\dfrac{99.\left(99+1\right)\left(2.99+1\right)}{6}=\dfrac{99.100.199}{6}=328350\)
TÍNH:
1/2.6 + 1/6.10 + 1/10.14 + ..... + 1/102.106
Đặt A = 1/2.6 + 1/6.10 + 1/10.14 + ..... + 1/102.106
=> 4A = 4/2.6 + 4/6.10 + 4/10.14 + ..... + 4/102.106
=> 4A = 1/2 - 1/6 + 1/6 - 1/10 + 1/10 - 1/14 + ... + 1/102 - 1/106
=> 4A = 1/2 - 1/106
=> 4A = 26/53
=> A = 13/106
~Study well~
#QASJ
\(\frac{1}{2.6}+\frac{1}{6.10}+...+\frac{1}{102.106}\)
\(=\frac{1}{4}.\left(\frac{4}{2.6}+\frac{4}{6.10}+...+\frac{4}{102.106}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+...+\frac{1}{102}-\frac{1}{106}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{106}\right)\)
\(=\frac{1}{4}.\frac{26}{53}\)
\(=\frac{13}{106}\)
Đặt \(B=\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{102.106}\)
\(\Leftrightarrow4B=\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{102.106}\)
\(\Leftrightarrow4B=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{102}-\frac{1}{106}\)
\(\Leftrightarrow4B=\frac{1}{2}-\frac{1}{106}=\frac{26}{53}\)
\(\Leftrightarrow B=\frac{26}{53}:4=\frac{13}{106}\)
Vậy ...................
~ Hok tốt ~
T= 2.6+6.10+10.14+.....+102.106
12.T=2.6.12+6.10.12+10.14.12+...+102.106.12=
=2.6.(10+2)+6.10.(14-2)+10.14.(18-6)+...+102.106.(110-98)=
=2.2.6+2.6.10-2.6.10+6.10.14-6.10.14+10.14.18-...-98.102.106+102.106.110=
=2.2.6+102.106.110
\(\Rightarrow T=\dfrac{2.2.6+102.106.110}{12}=99112\)
bạn j ơi cho mình hỏi là 102 lấy ở đâu ja <3
Tính
A=3/2.5+3/5.8+3/8.11+...+3/2015.2018
B=4/2.6+4/6.10+4/10.14+...+4/102.106
\(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2015.2018}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
Vậy \(A=\frac{504}{1009}.\)
\(B=\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{102.106}\)
\(=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{102}-\frac{1}{106}\)
\(=\frac{1}{2}-\frac{1}{106}=\frac{26}{53}\)
Vậy \(B=\frac{26}{53}.\)
Bài làm:
a) \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2015.2018}\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2018}\)
\(A=\frac{1}{2}-\frac{1}{2018}\)
\(A=\frac{504}{1009}\)
b) \(B=\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{102.106}\)
\(B=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{102}-\frac{1}{106}\)
\(B=\frac{1}{2}-\frac{1}{106}\)
\(B=\frac{26}{53}\)
\(A=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{2015.2018}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{2015}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
\(B=\frac{4}{2.6}+\frac{4}{6.10}+...+\frac{4}{102.106}=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+...+\frac{1}{102}-\frac{1}{106}=\frac{1}{2}-\frac{1}{106}\)
\(=\frac{26}{53}\)
D=3/2.6+3/6.10+3/10.14+...+3/24.30
Sửa đề: \(D=\dfrac{3}{2\cdot6}+\dfrac{3}{6\cdot10}+\dfrac{3}{10\cdot14}+...+\dfrac{3}{26\cdot30}\)
Ta có: \(D=\dfrac{3}{2\cdot6}+\dfrac{3}{6\cdot10}+\dfrac{3}{10\cdot14}+...+\dfrac{3}{26\cdot30}\)
\(=\dfrac{3}{4}\left(\dfrac{4}{2\cdot6}+\dfrac{4}{6\cdot10}+\dfrac{4}{10\cdot14}+...+\dfrac{4}{26\cdot30}\right)\)
\(=\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{26}-\dfrac{1}{30}\right)\)
\(=\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{30}\right)\)
\(=\dfrac{3}{4}\cdot\dfrac{28}{60}\)
\(=\dfrac{21}{60}=\dfrac{7}{20}\)
Câu 16: Viết số 946 dưới dạng tổng các lũy thừa của
10
là:
A.
946 9.100 40 6 = + + B.
3 2 1 946 9.10 4.10 6.10 = + +
C.
0 2 946 9.10 4.10 6.10 = + + D.
2 0 946 9.10 4.10 6.10 = + +