Tính tổng
F=1.2+3.4+5.6+6.7+........+97.98+99.100
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
3F = 1 . 2 . 3 + 3 . 4 . ( 5 - 2 ) + 5 . 6 . ( 7 - 4 ) +.....+ 99 . 100 . (101 - 98 )
3F = 1. 2 . 3 + 3. 4 . 5 - 2 . 3 . 4 + 5 . 6 . 7 - 4 . 5 . 6 +.....+ 99 . 100 . 101 - 98 . 99 . 100
3F = 1 . 2 . 3 + 99 . 100. 101
3F = 3 . 2 + 3 . 33 . 100 . 101
3F = 3 ( 2 + 333 300)
=>F = 3 . 333 302 : 3
=> F = 333 302
Vậy F = 333 302
G = \(1^2\)+\(2^2\)+ \(3^2\)+....+\(100^2\)
G=1 +2(1+1) +3(2+1) +..... + 100(99+1)
G=1 + 1.2+ 2 + 2.3 +3+ ......+ 99.100+100
G=(1+2+3+....+100) +(1.2+2.3+.....+99.100)
G= \(\frac{100\left(100+1\right)}{2}\)+\(\frac{100\left(100-1\right)\left(100+1\right)}{3}\)
G=5050+333300
G=338350
A=1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9
A=1/3-1/9
A=2/9
các câu 2;3 còn lại giống câu 1 bạn nhé
bạn thay số vào rồi làm tương tự
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Nhầm tưởng tính tích :v
Ta có :
\(B=\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}+\frac{1}{100}< \frac{1}{51}+\frac{1}{51}+...+\frac{1}{51}=50.\frac{1}{51}=\frac{50}{51}< \frac{99}{100}\)
\(\Leftrightarrow A>B\)
\(C=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{97.98}+\frac{1}{99.100}\)
\(C=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{99}-\frac{1}{100}\)
\(C=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{98}+\frac{1}{100}\right)\)
\(C=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(C=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)
\(C=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(D=\frac{1}{51.100}+\frac{1}{52.99}+\frac{1}{53.98}+...+\frac{1}{99.52}+\frac{1}{100.51}\)
\(D=\frac{1}{151}.\left(\frac{151}{51.100}+\frac{151}{52.99}+\frac{151}{53.98}+...+\frac{151}{99.52}+\frac{151}{100.51}\right)\)
\(D=\frac{1}{151}.\left(\frac{1}{100}+\frac{1}{51}+\frac{1}{99}+\frac{1}{52}+...+\frac{1}{52}+\frac{1}{99}+\frac{1}{51}+\frac{1}{100}\right)\)
\(D=\frac{1}{151}.\left(\frac{2}{100}+\frac{2}{99}+...+\frac{2}{51}\right)\)
\(D=\frac{2}{151}.\left(\frac{1}{100}+\frac{1}{99}+...+\frac{1}{51}\right)\)
\(\Rightarrow C:D=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{2}{151}.\left(\frac{1}{100}+\frac{1}{99}+...+\frac{1}{51}\right)}\)
\(\Rightarrow C:D=\frac{151}{2}=75\frac{1}{2}\)
Đặt A = 1.2 + 2.3 + 3.4 + ... + 99.100
3A = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ... + 99.100.(101-98)
3A = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
3A = 99.100.101
A = 33.100.101
A = 333300
\(A=1.2+2.3+3.4+4.5+...+97.98+98.99+99.100\)
\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+4.5.\left(6-3\right)+...+99.100.\left(101-98\right)\)
\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100\)
\(3A=99.100.101\)
\(A=\frac{99.100.101}{3}=\frac{999900}{3}=333300\)
Fx3=1x2x3+2x3x(4-1)+3x4x(5-2)+4x5x(6-3)+...+99x100x(101-98
Fx3=1x2x3+2x3x4-1x2x3+3x4x5-2x3x4+4x5x6-3x4x5+...+99x100x101-98x99x100
Fx3=99x100x101
F=333300