1x1=
2x2=
3x3=
4x4=
5x5=
mik mới lớp 2 nên mik chỉ bít đố thế thôi.
1x1=
2x2=
3x3=
4x4=
5x5=
1 x 1 = 1
2 x 2 = 4
3 x 3 = 9
4 x 4 = 16
5 x 5 = 25
Kết quả :
1 x 1 = 1
2 x 2 = 4
3 x 3 = 9
4 x 4 = 16
5 x 5 = 25
E= 1x1+2x2+3x3+4x4+...+99x99
G=1x100+2x99+3x98+...+99x2+100x1
giúp mik với mik đang vội
\(E=1.1+2.2+3.3+4.4+...+99.99\)
\(\Rightarrow E=1^2+2^2+3^2+4^2+...+99^2\)
\(\Rightarrow E=\dfrac{99.\left(99+1\right)\left(2.99+1\right)}{6}\)
\(\Rightarrow E=\dfrac{99.100.199}{6}\)
\(\Rightarrow E=33.50.199=328350\)
E = 1 x 1 + 2 x 2 + 3 x 3 + 4 x 4 +...+ 99 x 99
E = 1x(2-1) + 2 x (3-1)+...+ 99 x (100 -1)
D = 1 x 2 - 1 + 2 x 3 - 2 +...+ 99 x 100 - 99
D = 1x2 + 2 x 3 +...+ 99 x 100 - ( 1 + 2 +...+ 99)
Đặt A = 1x2 + 2 x 3 +...+ 99 x 100
B = 1 + 2 + ...+ 99
1x2 x 3 = 1x2x3
2x3x3 = 2x 3 x (4-1) = 2x3x4 - 1x2x3
3 x 4 x 3 = 3 x 4 x ( 5 - 2) = 3 x 4 x 5 - 2 x 3 x 4
................................................
99 x 100 x 3 = 99 x 100 x (101 - 98) = 99x100x101 - 98 x 99 x 100
Cộng vế với vế ta có: 3A = 99 x 100 x 101
A = 99 x 100 x 101 : 3 = 333300
B = 1 + 2 + 3 + ...+ 99
B = (99 + 1).[(99 -1):1 +1]:2 = 4950
E = 33300 - 4950 = 328350
a là cocaiconcac nhé e
b là cocaidaccau ok nhé
E=1x1+3x3+5x5+7x7+...+25x25
F=4x4+7x7+10x10+...+19x19
giúp mik với thank các bn
E= 2925
F= 951
nho k cho minh nha minh dang can diem
E= 1x1+3x3+5x5+7x7+9x9+11x11+13x13+15x15+17x17+19x19+21x21+25x25=2925
F= 4x4+7x7+10x10+13x13+16x16+19x19=951
bai nhu vay day cho minh xin mot k nha ko co cung duoc
1x1=1
2x2=4
3x3=9
4x4=?
5x5=?
1 x 1 = 1
2 x 2 = 4
3 x 3 = 9
4 x 4 = 16
5 x 5 = 25
t mk nha mọi người^^
Cho A=1+1x1!+2x2!+3x3!+4x4!+5x5!+...+10000x10000!
Tìm A dưới dạng n!
Ta có: \(n.n!=\left(n+1\right).n!-1.n!=\left(n+1\right)!-n!\)
Suy ra \(A=1+1.1!+2.2!+...+10000.10000!\)
\(=1+2!-1!+3!-2!+...+10001!-10000!\)
\(=10001!\)
1x1=
2x2=
3x3=
4x4=
5x5=
6x6=
7x7=
8x8=
9x9=
10x10=
1 x 1 = 1
2 x 2 = 4
3 x 3 = 9
4 x 4 = 16
5 x 5 = 25
6 x 6 = 36
7 x 7 = 49
8 x 8 = 64
9 x 9 = 81
10 x 10 = 100
\(1x1=1\)
\(2x2=4\)
\(3x3=9\)
\(4x4=16\)
\(5x5=25\)
\(6x6=36\)
\(7x7=49\)
\(8x8=64\)
\(9x9=81\)
\(10x10=100\)
1 x 1 = 1
2 x 2 = 4
3 x 3 = 9
4 x 4 = 16
5 x 5 = 25
6 x 6 = 36
7 x 7 = 49
8 x 8 = 64
9 x 9 = 81
10 x 10 = 100
Tính
A=1x2x3+2x3x3+3x4x3+4x5x3+....+98x99x3
B=1x2+2x3+3x4+4x5+...+98x99
C=1x1+2x2+3x3+4x4+5x5+...+98x98
A=1.2.3+2.3(4-1)+3.4(5-2)+4.5(6-3)+....+98.99(100-97) "." la dau nhan
A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+....+98.99.100-97.98.99
A=1.2.3+98.99.100
A= 970206
Ta có : B = 1.2 + 2.3 + 3.4 + ..... + 98.99
=> 3B = 0.1.2 + 1.2.3 - 1.2.3 + ...... + 98.99.100
=> 3B = 98.99.100
=> B = \(\frac{98.99.100}{3}\) = 323400
C = 1.1 + 2.2 + 3.3 + ..... + 98.98
=> C = 1.(2 - 1) + 2(3 - 1) + 3(4 - 1) + ...... + 98(99 - 1)
=> C = 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + ..... + 98.99 - 98
=> C = (1.2 + 2.3 + 3.4 + ..... + 98.99) - (1 + 2 + 3 + ..... + 98)
=> C = 323400 - 4851
=> C = 318549.
A=1x2+2x3+3x4+...+49x50
B=1x3+3x5+5x7+...+99x101
C=1x2x3+2x3x4+3x4x5+...+50x51x52
D=1x4+2x5+3x6+...+60x63
E=1x1+2x2+3x3+...+50x50
F=1x1+3x3+5x5+...+71x71
G=1x1+4x4+7x7+...+100x100
I=1x2x3+3x4x5+5x6x7+...+69x70x71
A=1x2+2x3+3x4+...+49x50
3A= 3(1.2+2.3+3.4+...+49.50)
3A= 1.2.3+2.3.3+3.4.3+...+49.50.3
3A= 1.2.(3-0)+2.3(4-1)+3.4(5-2)+...+49.50.(51-48)
3A= 0.1.2-1.2.3+1.2.3-2.3.4+2.3.4-3.4.5+...+48.49.50-49.50.51
3A= 49.50.51
A= 49.50.51/3=41650
B=1x3+3x5+5x7+...+99x101
B=1/1.3 +1/3.5 +...+1/99.101
2B=2/1.3 + 2/3.5 +...+2/99.101
2B=1-1/3+1/3-1/5+...+1/99-1/101
2B=1-1/101
2B=100/101
B=100/101:2=100/202
C=1x2x3+2x3x4+3x4x5+...+50x51x52
Nhân C với 4 ta được:
C x 4 = 1x2x3x4 + 2x3x4x 4 + 3x4x5x4 +…+50x51x52x4
C x 4 = 1x2x3x4 + 2x3x4x(5-1) + 3x4x5x(6-2) + ... + 50x51x52x(53-49)
C x 4 = 1x2x3x4 + 2x3x4x5 - 1x2x3x4 + 3x4x5x6 - 2x3x4x5 + ... +49x 50x51x52 - 50x51x52x53
Sau khi cộng - trừ giản ước ta có : C x 4 = 50x51x52x53
C = 50x51x52x53 : 4 = 1756950
S1=1x1
S2=2x2-1x1
S3=3x3-(2x2-1x1)
S4=4x4-(3x3-(2x2-1x1))
..................................
a)Hãy viết dãy số S5
b)Nếu dãy số này cứ tiếp tục như thế thì S2011 có giá trị là bao nhiêu?
S5=5x5-(4x4-(3x3-(2x2-1x1)))
S2011=2001x2001-(2000x2000-(1999x1999-(....)))
Bạn ơi tính như vậy thì phần b tính kiểu jr bao nhiêu dấu ngoạc làm sao tính được