D=1.4+2.5+3.6+...........+99.102
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\(C=1.4+2\left(4+1\right)+3\left(4+2\right)+...+99\left(4+98\right)\)
\(\Leftrightarrow C=1.4+2.4+1.2+3.4+2.3+...+99.4+98.99\)
\(C=\left(1.4+2.4+3.4+....+99.4\right)+\left(1.2+2.3+3.4+..98.99\right)\)
\(C=4\left(1+2+3+...+99\right)+\dfrac{1.2.3+2.3.3+3.4.3+...+98.99.3}{3}\)
\(C=\dfrac{4.\left(1+99\right).99}{2}+\dfrac{1.2.3+2.3\left(4-1\right)+...+98.99\left(100-97\right)}{3}\)
\(C=\dfrac{4.\left(1+99\right).99}{2}+\dfrac{1.2.3+2.3.4-1.2.3...+98.99.100-97.98.99}{3}\)
\(C=\dfrac{4.\left(1+99\right).99}{2}+\dfrac{98.99.100}{3}\)
\(C=19800+107800=127600\)
TK
S=1.4+2.5+3.6+4.7+....+n.(n+3) S = 1. ( 2 + 2 ) + 2. ( 3 + 2 ) + 3. ( 4 + 2 ) + . . . + n . [ ( n + 1 ) + 2 ] S = 1.2 + 2.3 + 3.4 + . . . . + n . ( n + 1 ) + ( 1.2 + 2.2 + 3.2 + . . . . + n .2 ) Đặt A = 1.2 + 2.3 + 3.4 + . . . . + n . ( n + 1 ) 3 A = 1.2.3 + 2.3. ( 4 − 1 ) + . . . . + n . ( n + 1 ) . [ ( n + 2 ) − ( n − 1 ) 3 A = 1.2.3 + 2.3.4 − 1.2.3 + . . . . + n . ( n + 1 ) . ( n + 2 ) − ( n − 1 ) . n . ( n + 1 ) 3 A = n . ( n + 1 ) . ( n + 2 ) A = [ n . ( n + 1 ) . ( n + 2 ) ] : 3 S = [ n . ( n + 1 ) . ( n + 2 ) ] : 3 + 2. ( 1 + 2 + 3 + . . . + n ) S = [ n . ( n + 1 ) . ( n + 2 ) ] : 3 + 2. n . ( n + 1 ) : 2 S = n . ( n + 1 ) . ( n + 2 ) : 3 + n . ( n + 1 ) S = n . ( n + 1 ) . [ ( n + 2 ) : 3 + 1 )
D = 1^2 + 2^2 + 3^2 + ... + n^2
= 1.( 2 - 1 ) + 2.( 3-1 ) + 3.( 4-1 ) + .... + n.[ ( n+ 1) - 1 ]
= 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + .... + n.( n+1 ) - n
= [ 1.2 + 2.3 + 3.4 + ..... + n.( n + 1 ) ] - ( 1 + 2 + 3 + .... + n )
= { [ n.( n+1 ).( n+2 )] /3 } - { [ n.( n+1)] /2 }
= { n(n+1)(2n+1) }/ 6
Vậy.........
D =1.4+2.5+3.6+.......+99.102
D = 1. (2+2) +2.(2+3) +3.(2+4)+...+99.(100+2)
D = 1.2+1.2+2.2+2.3+2.3+3.4+...+2.99+99.100
D = (1.2+2.3+3.4+...+99.100) +2.(1+2+3+4+...+99)
*Gọi A= 1.2+2.3+3.4+...+99.100
3A = 3.(1.2+2.3+3.4+...+99.100)
3A = 1.2.3+2.3.3+...+99.100.3
3A = 1.2.3 +2 .3.(4-1)+...+99.100.(101-98)
3A = 1.2.3+2.3.4-1.2.3+...+ 99.100.101-98.99.100
3A = 99.100.101
3A = 3.33.100.101
A = 33.100.101
A = 333300
* Gọi B = 2. (1+2+3+4+...+99)
\__có 99 số hạng ___/
B= 2.[(1+99).99:2]
B = 2 .4950
B = 9900
A+B = 333300+9900 =343200
Vậy D =343200
A = 1(2+2)+2(3+2)+3(4+2)+.+99(100+2)
A = 1.2+1.2+2.3+2.2+3.4+3.2+.+99.100+99.2
A = (1.2+2.3+3.4+.+99.100)+2(1+2+3+.+99)
Phần còn lại bạn tự động não đi nha.
C= (4/1.3).(9/2.4).(16/3.5)...........(10000/99.101)
C=(4.9.16...........10000)/(1.3).(2.4)......(99.101)
C=(2^2.3^2.4^2..........100^2)/(1.2.3.4......99).(3.4.5.......101)
C=(2.3.4........100).(2.3.4......100)/(1.2.3.....99).(3.4.5....101)
Sau khi triệt tiêu ở tử và mẫu ta được:
C=(2.100)/101
C=200/101
1.4+2.5+3.6+...+99.102=1(2+2)+2(3+2)+3(4+2)+...99(100+2)
=1.2+1.2+2.3+2.2+...+99.100+99.2
=(1.2+2.3+...+99.100)+2(1+2+...+99)
A=1.2+2.3+3.4+...+99.100(cho A la ten bieu thuc nay)
3A=1.2(3-0)+2.3(4-1)+...+99.100(101-98)
=(1.2.3+2.3.4+...+99.100.101)-(1.2.3+2.3.4+3.4.5+...+98.99.100
=99.100.101=>A=\(\frac{99.100.101}{3}\)=33330
2.(1+2+...99)
=2(100.99:2)=2.4950=9900
33330+9900=343200
vay...
Đặt \(A=1.4+2.5+3.6+...+100.103\)
\(=1\left(2.2\right)+2\left(3+2\right)+3\left(4+2\right)+...+100\left(101+2\right)\)
\(=1.2+2.3+3.4+...+100.101+\left(1.2+2.2+3.2+...+100.2\right)\)
\(=1.2+2.3+3.4+...+100.101+2\left(1+2+3+...+100\right)\)
\(=1.2+2.3+3.4+...+100.101+2.100\left(100+1\right):2\)
\(=1.2+2.3+3.4+...+100.101+10100\)
Đặt \(B=1.2+2.3+3.4+...+100.101\)
\(\Rightarrow3B=1.2.3+2.3.3+3.4.3+100.101.3\)
\(\Rightarrow3B=1.2.3+2.3\left(4-1\right)+3.4\left(5-2\right)+...+100.101\left(102-99\right)\)
\(\Rightarrow3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+100.101.102-99.100.101\)
\(\Rightarrow3B=100.101.102\)
\(\Rightarrow B=343400\)
Khi đó \(A=343400=10100=333300\)
Đặt A = 1.4 + 2.5 + 3.6 + 4.7 + ... + 100.103
3A = 3.(1.2 + 2.3 + 3.4 + ... + 100.101] + 3.(2 + 4 + 6 + ... + 200)
= 1.2.3 + 2.3.3 + 3.4.3 + ... + 100.101.3 + 3.(2 + 4 + 6 + ... + 200)
\(\Rightarrow\) A = 100.101.105:3 = 353500
3 . 6 = 3 . 4 + 2 . 3 rùi đấy bạn, bn xét từng tích rùi sẽ thấy thôi.
Baig iair