1/1+1/2+1/3+...+1/30
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LM
0
3 tháng 2 2022
a) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
=\(1-\dfrac{1}{6}\)=\(\dfrac{5}{6}\)
b) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
=\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
=\(\dfrac{1.2}{3.5.2}+\dfrac{1.2}{5.7.2}+\dfrac{1.2}{7.9.2}+\dfrac{1.2}{9.11.2}+\dfrac{1.2}{11.13.2}\)
=\(\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\).
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)=\(\dfrac{1}{2}.\dfrac{10}{39}\)=\(\dfrac{5}{39}\).
c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
=\(1-\dfrac{1}{8}=\dfrac{7}{8}\).
d) \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4}{2^5}+\dfrac{2^3}{2^5}+\dfrac{2^2}{2^5}+\dfrac{2}{2^5}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4+2^3+2^2+2+1}{2^5}\)=\(\dfrac{2^5-1}{2^5}=\dfrac{31}{32}\).
e) \(\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}=\dfrac{7^{99}+7^{98}+7^{97}+...+7+1}{7^{100}}=\dfrac{\dfrac{7^{100}-1}{6}}{7^{100}}=\dfrac{7^{100}-1}{6.7^{100}}\)
16 tháng 11 2020
B=1+1/2*(1+2)*2/2+1/3*(1+3)*3/2+....+1/30*(1+30)*30/2
=1+1+2/2+1+3/2+...+1+30/2
=2+3+4+...+31/2
=(31+2)*30/2 phần 2
=) B=247,5
NP
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 9 2023
A = 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4)+...+ (1 + 2 + 3+...+ 30)
A = 1 + (2+1)\(\times\)2: 2+ (3+1)\(\times\)3: 2 + ....+ (30 +1) \(\times\) 30: 2
A = 1 + 2 \(\times\)3: 2 + 3 \(\times\) 4 : 2 +...+ 30 \(\times\) 31:2
A = \(\dfrac{2+2\times3+3\times4+...+30\times31}{2}\)
Đặt B = 2 + 2 \(\times\) 3 + 3 \(\times\) 4 +...+ 30 \(\times\) 31
1 x 2 x 3 = 1 x 2 x3 = 1 x 2 x 3
2 x 3 x 3 = 2 x 3 x ( 4 - 1) = 2 x 3 x 4 - 1 x 2 x 3
3 x 4 x 3 = 3 x 4 x (5- 2) = 3 x 4 x 5 - 2 x 3 x 4
...........................................................................
30 x 31 x 3 = 30 x 31 x (32 - 29) = 30 x 31 x 32 - 29 x 31 x 30
Cộng vế với vế ta có:
B x 3 = 30 x 31 x 32 ⇒ B = 30 x 31 x 32 : 3 = 14880
A = B : 2 = 14880 : 2 = 7440
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
22 tháng 3
Câu 11:
(\(\dfrac{11}{4}\). \(\dfrac{-5}{9}\) - \(\dfrac{4}{9}\).\(\dfrac{11}{4}\)).\(\dfrac{8}{33}\)
= \(\dfrac{11}{4}\).(\(\dfrac{-5}{9}\) - \(\dfrac{4}{9}\)). \(\dfrac{8}{33}\)
= \(\dfrac{11}{4}\).(-1).\(\dfrac{8}{33}\)
= - \(\dfrac{2}{3}\)
HS
1
20 tháng 6 2021
ĐKXĐ: x khác 0
1/(1 ×2)+ 1/(2 ×3)+ 1/(3 ×4 )+⋯.+ 1/(x ×( x+1))= 30/31
1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/x - 1/x +1 = 30/31
1 - 1/x+1 = 30/31
x+1-1/x+1 = 30/31
x/x+1 = 30/31
31x = 30x + 30
31x - 30x = 30
x = 30(TM)
Vậy x là 30.