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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
=> 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
đúng nha
Ta có:a=1.2+2.3+3.4+...+98.99
a=1(1+1)+2(2+1)+3(3+1)+...+98(98+1)
a=12+1+22+2+33+3+...+982+98
a=b+(1+2+3+4+...+98)
a=b+(1+98).98:2
a=b+4851
Vậy a-b =4851
Bài giải
\(B=1\cdot2^2+2\cdot3^2+3\cdot4^2+...+99\cdot100^2\)
\(B=1\cdot2\cdot\left(3-1\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-1\right)+...+99\cdot100\cdot\left(101-1\right)\)
\(B=1\cdot2\cdot3-1\cdot2+2\cdot3\cdot4-2\cdot3+...+99\cdot100\cdot101-99\cdot100\)
\(B=\left(1\cdot2\cdot3+2\cdot3\cdot4+...+99\cdot100\cdot101\right)-\left(1\cdot2+2\cdot3+...+99\cdot100\right)\)
Đặt \(C=1\cdot2\cdot3+2\cdot3\cdot4+...+99\cdot100\cdot101\)
\(4C=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+...+99\cdot100\cdot101\cdot\left(102-98\right)\)
\(4C=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+...+99\cdot100\cdot101\cdot102-98\cdot99\cdot100\cdot101\)
\(4C=99\cdot100\cdot101\cdot102\)
\(4C=101989800\)
\(C=101989800\text{ : }4\)
\(C=25497450\)
* Khai triển
1.2^2 = 1.2.2 = 1.2.(3 - 1) = 1.2.3 - 1.2
2.3^2 = 2.3.3 = 2.3.(4 - 1) = 2.3.4 - 2.3
3.4^2 = 3.4.4 = 3.4(5 - 1) = 3.4.5 - 3.4
.....................................................
98.99^2 = 98.99.99 = 98.99.100 - 98.99
Vậy
E = 1.2.3+2.3.4 + 3.4.5 + ... + 98.99.100 - (1.2 + 2.3 + 3.4 + ..+ 98.99) = X - Y
Ta có
X = 1.2.3+2.3.4 + 3.4.5 + ... + 98.99.100
X.4 = 1.2.3.4 + 2.3.4.(5 - 1) + 3.4.5.(6 - 2) +....+98.99.100.(101-97) = 98.99.100.101
=> X = 98.99.100.101/4 = ....
Y = 1.2 + 2.3 + 3.4 + ..+ 98.99
Y.3 = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + .. + 98.99.(100-97) = 98.99.100
=> Y = 98.99.100/3 = ...
Vậy E = X - Y = .... - .... = 24174150
Chứng tỏ rằng :
a) 1 phần 1.2 + 1 phần 2.3 + 1 phần 3.4+.....+1 phần 49.50 <1
b)1 phần 22 + 1 phần 32 + 1 phần 42+.....+1 phần 20082 + 1 phần 20092 <1
Toán lớp 6
ai tích mình tích lại
Khỏi cần tích.
A=1.22+2.32+...+2017.20182
=(2-1).22+(3-1).32+...+(2018-1).20182
=23+33+43+...+20183 - (22+32+...+20182)
=(13+23+...+20183)-(12+22+...+20182)
Ta có: 4k3=(k4+2k3+k2)-(k4-2k3+k2)=[(k+1)k]2-[k(k-1)]2
=>
4.20183=(2019.2018)2-(2018.2017)2
4.20173=(2018.2017)2-(2017.2016)2
...
4.23=(3.2)2-(2.1)2
4.13=(2.1)2-(1.0)2
Cộng 2 vế:
4(13+23+...+20183)=(2019.2018)2 => 13+23+...+20183=(2019.2018)2 /4
Lại có:
6.k2=[2(k+1)3-3(k+1)2+(k+1)] - [2.k3-3k2+k]
=>
6.20182=(2.20193-3.20192+2019)-(2.20183-3.20182+2018)
6.20172=(2.20183-3.20182+2018)-(2.20173-3.20172+2017)
....
6.22=(2.33-3.32+3)-(2.23-3.22+2)
6.12=(2.23-3.22+2)-(2.13-3.12+1)
Cộng 2 vế:
6(12+22+...+20182)=(2.20193-3.20192+2019) =>12+22+...+20182=(2.20193-3.20192+2019)/6
=>
A=[(2019.2018)2 /4] - [(2.20193-3.20192+2019)/6]