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E = 1.2+2.3+3.4+......+99.100
Gấp E lên 3 lần ta có:
E . 3 = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3
E . 3 = 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98)
E . 3 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100 E . 3 = 99.100.101
E = 99.100.101 : 3
E = 33.100.101
E = 333 300
k mik nha
E = 1.2 + 2.3 + 3.4 + ... + 99.100
=> 3E = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3
=> 3E = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) +...+ 99.100.(101-98)
=> 3E = 1.2.3 - 0 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
=> 3E = 99.100.101
=> E = 333300
TL:
a)\(2+4+6+...+2000=\frac{\left(2+2000\right).\left[\left(2000-2\right):2+1\right]}{2}\)
\(=1001000\)
Câu b tương tự nha bạn:)
c) Đặt 1.2+2.3+....+99.100 =A
\(3A=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)\)
\(3A=1.2.3+2.3.4-1.2.3+...99.100.101-98.99.100\)
\(3A=99.100.101\)
\(A=333300\)
Vậy .....
a) Đặt A= 2+4+6+...+1998+2000
Ta có: A=(2+2000).1000:2
=> A=2002.1000:2
=> A=2002000:2
=> A=1001000
b) Đặt B= 5+9+13+...+1997+2001
=> B=(2001+5).500:2
=> B=2006.500:2
=> B=1003000:2
=> B=501500
c)Đặt S= 1.2 + 2.3 + 3.4 + ...+ 99.100
=> 3S = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3
3S= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)
3S= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100
3S = 99.100.101 => 3S = 3.33.100.101
=> S=33.100.101= 333300
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(=1+0+0+0+0+0-\dfrac{1}{7}\)
\(=1-\dfrac{1}{7}\)
\(=\dfrac{7}{7}-\dfrac{1}{7}\)
\(=\dfrac{6}{7}\)
\(\text{Ta có: A = 1.2+2.3+3.4+4.5+...+99.100 }\)
=> 3A = 3.(1.2+2.3+3.4+4.5+...+99.100)
=> 3A = 1.2.(3 - 0) +2.3.(4 - 1) + 3.4.(5-2) + ........ + 99.100.(101 - 98)
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .......... + 99.100.101
=> 3A = 99.100.101
\(\Rightarrow A=\frac{99.100.101}{3}=333300\)
k mình nếu đúng OK
Áp dụng công thức ta có :
\(A=1.2+2.3+3.4+...+99.100=\frac{99.100.101}{3}=333300\)
A = 1.2 + 2.3 + 3.4 + ....... + 99.100
3A = 1.2.3 + 2.3.3 + 3.4.3 + ....... + 99 . 100 . 3
3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) +.... + 99.100.(101-98)
3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ..... + 99 . 100 . 101 - 98 . 99 . 100
3A = (1.2.3 - 1.2.3) + (2.3.4-2.3.4) + ... + (98.99.100 - 98.99.100) + 99 . 100 . 101
3A = 99 . 100 . 101 = 999900
A = 999900 : 3 = 333300
A=1*2+2*3+3*4+...+99*100
A=100*101*102:3
A=343400(công thức)
gọi tổng là S ta có
3S=1.2.3-0.1.2+2.3.4-1.2.3+......+99.100.101-98.99.100
=>3S=98.99.100
=>S=\(\frac{98.99.100}{3}=323400\)
S=1.2+ 2.3+4,5.......+99.100
Nhân cả 2 vế với 3, ta được:
3S=1.2.3+ 2.3.3+ 3.4.3+ 4.5.3+...... 99.100.3
= 1.2.3 + 2.3(4-1) + 3.4.(5-2) +...+ 99.100.(101-98)
= 1.2.3 + 2.3.4 -1.2.3 + 3.4.5-2.3.4 +...+ 99.100.101-98.99.100
= 99.100.101
----> S = (99.100.101):3
S= 333300
Vậy A=333300
A = 1.2 + 2.3 + 3.4 + ... + 99.100
3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3
3A = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) +...+ 99.100.(101-98)
3A = 1.2.3 - 0 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
3A = 99.100.101
A = 333300