Tính : 1.3 + 2.4 + 3.6 + ..... + 98.100 + 99.101
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GD
0
8 tháng 10 2019
A = 1×3+3×5+5×7+...+ 97×99+99×101
6A= 1×3×6+3×5×6+5×7×6+...+97×99×6+99×101×6
6A= 1×3×(5+1)+3×5×(7-1)+5×7×(9-3)+...+97×99×(101-95)+99×101×(103-97)
6A = 1×3×5-1×3+3×5×7-1×3×5+5×7×9-3×5×7+7×9×11-5×7×9+,,,+97×99×101-95×97×99+99×101×103-97×99×101
6A= 1×3+99×101×103
6A= 1029900
A= 171650
NK
0
TK
0
10 giờ trước (12:35)
Ta có: \(A=1\cdot99+2\cdot98+3\cdot97+\cdots+98\cdot2+99\cdot1\)
\(=2\left(1\cdot99+2\cdot98+\cdots+49\cdot51\right)+50\cdot50\)
\(=2\left\lbrack1\left(100-1\right)+2\left(100-2\right)+\cdots+49\left(100-49\right)\right\rbrack+2500\)
\(=2\cdot\left\lbrack100\left(1+2+\cdots+49\right)-\left(1^2+2^2+\cdots+49^2\right)\right\rbrack+2500\)
\(=2\cdot\left\lbrack100\cdot\frac{49\cdot50}{2}-\frac{49\cdot\left(49+1\right)\left(2\cdot49+1\right)}{6}\right\rbrack+2500\)
\(=2\left\lbrack50\cdot49\cdot50-\frac{49\cdot50\cdot99}{6}\right\rbrack+2500\)
\(=2\cdot\left\lbrack49\cdot50\cdot50-49\cdot25\cdot33\right\rbrack+2500\)
\(=2\cdot49\cdot25\cdot\left(2\cdot50-33\right)+2500\)
\(=49\cdot50\cdot67+2500=166650\)
Ta có: \(B=1\cdot2\cdot3+2\cdot3\cdot4+\ldots+17\cdot18\cdot19\)
\(=2\left(2-1\right)\left(2+1\right)+3\left(3-1\right)\left(3+1\right)+\cdots+18\left(18-1\right)\left(18+1\right)\)
\(=2\cdot\left(2^2-1\right)+3\left(3^2-1\right)+\cdots+18\left(18^2-1\right)\)
\(=\left(2^3+3^3+\cdots+18^3\right)-\left(2+3+\cdots+18\right)\)
\(=\left(1^3+2^3+\cdots+18^3\right)-\left(1+2+3+\cdots+18\right)\)
\(=\left(1+2+\cdots+18\right)^2-\left(1+2+\cdots+18\right)\)
\(=\left(18\cdot\frac{19}{2}\right)^2-18\cdot\frac{19}{2}=\left(9\cdot19\right)^2-9\cdot19=29070\)
Ta có: \(C=1\cdot4+2\cdot5+\cdots+100\cdot103\)
\(=1\left(1+3\right)+2\left(2+3\right)+\cdots+100\cdot\left(100+3\right)\)
\(=\left(1^2+2^2+\cdots+100^2\right)+3\left(1+2+\cdots+100\right)\)
\(=\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}+\frac{3\cdot100\cdot101}{2}\)
\(=\frac{100\cdot101\cdot201}{6}+\frac{3\cdot100\cdot101}{2}=50\cdot101\cdot67+3\cdot50\cdot101\)
\(=50\cdot101\cdot70=3500\cdot101=353500\)
Ta có: \(D=1\cdot3+2\cdot4+3\cdot5+\cdots+97\cdot99+98\cdot100\)
\(=1\left(1+2\right)+2\left(2+2\right)+3\left(3+2\right)+\cdots+97\cdot\left(97+2\right)+98\cdot\left(98+2\right)\)
\(=\left(1^2+2^2+\cdots+98^2\right)+2\cdot\left(1+2+3+\cdots+98\right)\)
\(=\frac{98\cdot\left(98+1\right)\left(2\cdot98+1\right)}{6}+2\cdot\frac{98\cdot99}{2}\)
\(=\frac{98\cdot99\cdot197}{6}+98\cdot99=49\cdot33\cdot197+98\cdot99=49\cdot33\left(197+2\cdot3\right)\)
\(=49\cdot33\cdot203=328251\)
HL
26 tháng 7 2019
\(A=1.\left(2+2\right)+2.\left(3+2\right)+3.\left(4+2\right)+....+99.\left(100+2\right)\)
\(A = (1.2 + 2.3 + 3.4 + ... + 99.100) + (1.2 + 2.2 + 3.2 + ... + 99.2)\)
\(Đặt B = 1.2 + 2.3 + 3.4 + ... + 99.100\)
\(3B = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ... + 99.100.(101-98)\)
\(3B = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100\)
\(3B = 99.100.101\)
\(B = 33.100.101 = 333300\)
\(A = 333300 + 2.(1 + 2 + 3 + ... + 99)\)
\(A = 333300 + 2.(1 + 99).99:2\)
\(A = 333300 + 100.99\)
\(A = 333300 + 9900\)
\(A = 343200\)
CC
26 tháng 7 2019
a. A = 1.4 + 2.5 + 3.6 +...+ 99.102
= 1( 2 +2) + 2(3+2) +...+ 99 (100 +2)
= 1.2 + 1.2 +2.3 + 2.2 +...+ 99 .100 +99 . 2
= ( 1.2 +2.3 + 3.4 +...+99 . 100) + 2(1 + 2 + 3+...+99)
= 333300 + 9900 = 343 200
b. B = 1.3 + 2.4 + 3.5 +...+ 99.101
= 1(2 +1) + 2(3 +1) + 3(4 +1) +...+ 99(100 +1)
= 1.2 + 1 + 2.3 + 2 + 3.4 +...+ 99. 100 +99
= ( 1.2 + 2.3 + 3.4 +...+ 99.100) + (1+2+...+99)
= 333300 + 4950 = 338 250
c. C = 4 + 12 + 24 +...+ 19404 + 19800
1/2C = 1.2 + 2.3 + 3.4 +...+ 98.99 + 99.100
1/2 C = 333300
C = 333300 : 1/2 = 666600
đặt A=1.3 + 2.4 + 3.6 + ..... + 98.100 + 99.101
=1.(2+1)+2(3+1)+...+99(100+1)
=1.2+1+2.3+2+...+99.100+99
=(1.2+2.3+...+99.100)+(1+2+...+99)
đặt B =(1.2+2.3+...+99.100)
B= (99.100.101-0.1.2):3=333300
đặt C = (1+2+...+99)
SSH C:(99-1)x1+1=99
tổng C=(99+1)x99:2=4950
ta có A=B+C
=> A= 333300 + 4950 = 338250
=> 1.3 + 2.4 + 3.6 + ..... + 98.100 + 99.101=338250
=(99.101.103-0.1.3):3
=....... ( tự điền nhé)