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\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(\frac{1}{3.5.}\right).....\left(1+\frac{1}{99.101}\right)\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.....\frac{10000}{9999}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{100^2}{99.101}\)
\(=\frac{2^2.3^2.4^2.5^2.....98^2.99^2.100^2}{1.2.3^2.4^2.5^2......99^2.100.101}\)
\(=\frac{2.100}{1.101}\)
\(=\frac{200}{101}\)
\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)...\left(1+\frac{1}{99.101}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}....\frac{10000}{99.101}\)
\(=\frac{2.2.3.3...100.100}{1.3.2.4...99.101}\)
\(=\frac{\left(2.3.4...100\right)\left(2.3.4...100\right)}{\left(1.2...99\right)\left(3.4.5...101\right)}\)
\(=\frac{100.2}{101}=\frac{200}{101}\)
\(D=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
\(D=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{10000}{99.101}\)
\(D=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{100^2}{99.101}\)
\(D=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4.5...101}=100.\frac{2}{101}=\frac{200}{101}\)
Vậy \(D=\frac{200}{101}\)
\(A=1+2+2^2+...+2^{2018}\)
\(2A=2+2^2+...+2^{2019}\)
\(2A-A=\left[2+2^2+...+2^{2019}\right]-\left[1+2+2^2+...+2^{2018}\right]\)
\(A=2^{2019}-1\)
#)Giải :
\(A=1+2+2^2+2^3+...+2^{2018}\)
\(2A=2+2^2+2^3+2^4+...+2^{2019}\)
\(2A-A=\left(2+2^2+2^3+2^4+...+2^{2019}\right)-\left(1+2+2^2+2^3+...+2^{2018}\right)\)
\(A=2^{2019}-1\)
\(B=3+3^2+3^3+...+3^{2017}\)
\(3B=3^2+3^3+3^4+...+3^{2018}\)
\(3B-B=\left(3^2+3^3+3^4+...+3^{2018}\right)-\left(3+3^2+3^3+...+3^{2017}\right)\)
\(2B=3^{2018}-3\)
\(B=\frac{3^{2018}-3}{2}\)
Xét số hạng tổng quát:
1 + 1/[k.(k + 2)] = [k.(k + 2) + 1]/[k.(k + 2)] = (k + 1)²/[k.(k + 1)], với k nguyên dương.
Cho k chạy từ 1 đến 99, ta có:
• 1 + 1/1.3 = 2²/(1.3).
• 1 + 1/2.4 = 3²/(2.4).
• 1 + 1/3.5 = 4²/(3.5).
.......................
• 1 + 1/97.99 = 98²/(97.99).
• 1 + 1/98.100 = 99²/(98.100).
• 1 + 1/99.101 = 100²/(99.101).
Nhân vế với vế các đẳng thức trên, ta được:
(1 + 1/1.3).(1 + 1/2.4)(1 + 1/3.5)....(1 + 1/99.101)
= [2².3².....100²]/[1.2.3².4²......99².100...
= (2².100²)/(2.100.101)
= 2.100/101
= 200/101.
còn N thì chịu
M=(4/1.3.9/2.4.16/3.5...10000/99.101
M=2.2/1.3.3.3/2.4.4.4/3.5...100.100/99.101
M=2.3.4.5...100/1.2.3...99.3.4.5...100/2.3.4.5...101
M=100.2/101=200/101
Cau N sai de rui ban a, o mau so phai la 1.5.7+2.10.14+4.20.28+7.35.49 moi lam dc.
