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HL
1
31 tháng 1
\(A=1\cdot2+2\cdot3+...+151\cdot152\)
\(=1\left(1+1\right)+2\left(1+2\right)+...+151\left(1+151\right)\)
\(=\left(1+2+3+...+151\right)+\left(1^2+2^2+...+151^2\right)\)
\(=\dfrac{151\left(151+1\right)}{2}+\dfrac{151\left(151+1\right)\left(2\cdot151+1\right)}{6}\)
\(=151\cdot76+\dfrac{151\cdot152\cdot303}{6}\)
\(=151\cdot76+151\cdot7676=1170552\)
\(C=2\cdot4+4\cdot6+...+2024\cdot2026\)
\(=2\cdot2\left(1\cdot2+2\cdot3+...+1012\cdot1013\right)\)
\(=4\left[1\left(1+1\right)+2\left(1+2\right)+...+1012\left(1+1012\right)\right]\)
\(=4\left[\left(1+2+...+1012\right)+\left(1^2+2^2+...+1012^2\right)\right]\)
\(=4\left[1012\cdot\dfrac{1013}{2}+\dfrac{1012\left(1012+1\right)\left(2\cdot1012+1\right)}{6}\right]\)
\(=4\left[506\cdot1013+345990150\right]\)
\(=1386010912\)
\(M=1^2+2^2+...+2024^2\)
\(=\dfrac{2024\left(2024+1\right)\cdot\left(2\cdot2024+1\right)}{6}\)
\(=2024\cdot2025\cdot\dfrac{4049}{6}\)
=2765871900
\(N=1^3+2^3+...+100^3\)
\(=\left(1+2+3+...+100\right)^2\)
\(=\left[\dfrac{100\left(100+1\right)}{2}\right]^2\)
\(=\left[50\cdot101\right]^2=5050^2\)
\(Q=1^3+2^3+...+2024^3\)
\(=\left(1+2+3+...+2024\right)^2\)
\(=\left[\dfrac{2024\left(2024+1\right)}{2}\right]^2\)
\(=\left[1012\left(2024+1\right)\right]^2\)
\(=2049300^2\)
TT
1
25 tháng 8 2017
Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 32.33
=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 32.33.34
=> 3S = 32.33.34
=> S = \(\frac{32.33.34}{3}=11968\)
6 tháng 7 2019
a) Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32
A = 1/2 + 1/22 + 1/23 + 1/24 + 1/25
2A = 2(1/2 + 22 + 1/23 + 1/24 + 1/25)
2A = 1 + 1/2 + 1/22 + 1/23 + 1/24
2A - A = (1 + 1/2 + 1/22 + 1/23 + 1/24) - (1/2 + 1/22 + 1/23 + 1/24 + 1/25)
A = 1 - 1/25
A = 31/32
b) 2/1.2 + 2/2.3 + 2/3.4 + ... + 2/18 . 19 + 2/19.20
= 2(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/18.19 + 1/19.20)
= 2.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/18 - 1/19 + 1/19 - 1/20)
= 2. (1 - 1/20)
= 2.19/20
= 19/10
KG
Cho A=1/1.2 + 1/2.3 + + 1/ 3.4+...+1/49.50 ; B = 1.2+2.3+3.4+4.5+5.6+...+49.50
Tính 50 mủ 2 A – B/17
0
TT
1
12 tháng 3 2018
BÀI 1:
\(S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(S=1+\frac{1}{1.2}+\frac{1}{2.2}+\frac{1}{2.4}+\frac{1}{4.4}+\frac{1}{4.8}\)
\(S=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}\)
\(S=1+1-\frac{1}{8}\)
\(S=\frac{15}{8}\)
BÀI 2:
\(A=1.2+2.3+3.4+...+98.99\)
\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+98.99.3\)
\(3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+98.99.\left(100-97\right)\)
\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.99\)
\(3A=\left(1.2.3+2.3.4+3.4.5+98.99.100\right)-\left(1.2.3+2.3.4+...+97.98.99\right)\)
\(3A=98.99.100\)
\(3A=970200\)
\(\Rightarrow A=970200:3\)
\(A=323400\)
CHÚC BN HỌC TỐT!!!
DD
Đoàn Đức Hà
Giáo viên
12 tháng 7 2021
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(B=1.2+2.3+3.4+...+49.50\)
\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)
\(=49.50.51\)
\(B=\frac{49.50.51}{3}=49.50.17\)
\(50^2.A-\frac{B}{17}=49.50-49.50=0\)
S
5
31 tháng 1 2016
a ta co ;
13 -12 +11+10-9+8-7-6+5-4+3+2-1
=13-(12-11-10+9) +(8-7-6+5) -(4-3-2+1)
= 13 -0+0 -0
=13
sorry , ko để ý,
Ta có A =1.2 + 2.3 + 3.4 + ...+ 98.99
B = 1^2 + 2^2 + 3^2 +...+98^2 = 1.1+2.2+3.3+...+98.98
Suy ra: A-B= (1.2 + 2.3 + 3.4 + ...+ 98.99) - (1.1+2.2+3.3+...+98.98)
= (1.2-1.1) + (2.3-2.2) + (3.4-3.3) +...+ (98.99-98.98)
= 1(2-1) + 2(3-2) + 3(4-3) +...+ 98(99-98)
= 1.1 + 2.1 + 3.1 +...+ 98.1
= 1+ 2+ 3+...+ 98 = [98.(98+1)]/2= 98.99/2 = 4851
A = 1.2 + 2.3 + 3.4 + ... + 98.99
A x 3 =1.2.3 + 2.3.3 + 3.4.3 + ... + 98.99.3
A x 3 = 1.2.3 + 2.3.(4-1 ) + 3.4.(5-2 )+...+98.99.(100-97)
A x 3 = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.99
A x 3 = 98.99.100
=> A = 98.99.100:3
=> A = 323400