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G= \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+...+98.99}\)
G= \(\frac{\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+...+\frac{98.99}{2}}{1.2+2.3+3.4+...+98.99}\)
G = \(\frac{\frac{1.2+2.3+...+98.99}{2}}{1.2+2.3+3.4+...+98.99}\)
G= \(\frac{1}{2}\)
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 + 32 + 3 + ... + 982.98
A = (12 + 22 + 32 + ... + 982) + (1 + 2 + 3 + ... + 98)
A = (12 + 22 + 32 + ... + 982) + 4851 (1)
B = 12 + 22 + 32 + ... + 982 (2)
(1)(2) => A - B = 4851 ⋮ 4851
ta có: B = 12 + 22 + 32 +...+982 = 1.1 +2.2+3.3+...+98.98
=> A-B = (1.2+2.3+3.4+4.5+...+98.99) - (1.1+2.2+3.3+...+98.98)
A-B = (1.2-1.1) + (2.3-2.2) + (3.4-3.3) + (4.5-4.4) + ...+ (98.99-98.98)
A-B = 1.(2-1) + 2.(3-2) +3.(4-3) + 4.(5-4) + ...+ 98.(99-98)
A-B = 1 +2+3+4+...+98
A-B = (1+98).98:2
A -B = 4851 chia hết cho 4851
Câu hỏi của Nguyễn Hồ Yến Ngân - Toán lớp 6 - Học toán với OnlineMath
Em tham khảo nhé!
Đặt tổng trên là A , ta có :
\(\frac{A}{2}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(\frac{A}{2}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(\frac{A}{2}=\left(1-\frac{1}{100}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...+\left(\frac{1}{98}-\frac{1}{98}\right)+\left(\frac{1}{99}-\frac{1}{99}\right)\)\(\frac{A}{2}=\frac{99}{100}\)
\(A=\frac{99}{100}.2\)
\(A=\frac{99}{50}\)
\(S=\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{98\times99}+\frac{2}{99\times100}\)
\(S=2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\right)\)
\(S=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(S=2\times\left(1-\frac{1}{100}\right)\)
\(S=2\times\frac{99}{100}\)
\(S=\frac{99}{50}\)
\(S=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{98.99}+\frac{2}{99.100}\)
\(S=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(S=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}+\frac{1}{100}\right)\)
\(S=2.\left(\frac{1}{1}-\frac{1}{100}\right)\\ S=2.\left(\frac{100}{100}+\frac{-1}{100}\right)\\ S=2.\frac{99}{100}\\ S=\frac{99}{50}\)
a-b=(1.2+2.3+3.4+4.5+...+98.99)-(12+22+32+...+982)
1.2+2.3+3.4+4.5+...+98.99-12-22-32-...-982
=1(2-1)+2(3-2)+...+98(99-98)
=1+2+...+98
Đến đây bạn tự tính