S = 2/2.3 + 2/3.4 + 2/4.5 +...+ 2/59.60 . Chứng tỏ S < 1.
Tìm S2 = 1.2+2.3+3.4+...+n.(n+1)
tính tổng S= (1.2)² + (2.3)² + (3.4)² + … + [n(n + 1)]²
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Bài 1:
a, CMR: A = \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{21}{10^2.11^2}< 1\)
b, Cho B = \(\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+\dfrac{24}{25}+...+\dfrac{2499}{2500}.\) CMR: B không phải là số nguyên.
c, So sánh: C = \(\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{2021}{2^{2020}}\) với 3.
Chứng minh zới k thuộc N* ta luôn có: k(k+1)(k+2)-(k-1)k(k+1) áp dụng tính tổng: S=1.2+2.3+3.4+..............+n(n+1)
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\(\text{Tìm x, biết:}\)
\(a\)) \(x-\dfrac{2}{3.5}-\dfrac{2}{5.7}-\dfrac{2}{7.9}-\dfrac{2}{9.11}-\dfrac{2}{11.13}-\dfrac{2}{13.15}=\dfrac{2}{5}\)
\(b\)) \(\dfrac{1}{2.3}.x+\dfrac{1}{3.4}.x+\dfrac{1}{4.5}.x+...+\dfrac{1}{49.50}.x=1\)
\(c\)) \(x-\dfrac{20}{11.3}-\dfrac{20}{13.15}-...-\dfrac{53}{55}=\dfrac{3}{11}\)
\(d\)) \(x+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{41.45}=\dfrac{-37}{45}\)
\(e\)) \(\left(\dfrac{11}{12}.\dfrac{11}{2.23}.\dfrac{11}{23.34}...\dfrac{11}{89.100}\right).x=\dfrac{5}{3}\)
\(f\)) \(\left(\dfrac{2}{11.13}.\dfrac{2}{13.15}.\dfrac{2}{15.17}...\dfrac{2}{19.21}\right)-x+4+\dfrac{221}{231}=\dfrac{7}{3}\)
Cho M=2018 +20182+20183+...+20182018
CMR M chia het cho 2019
S2M Voi N=22019/2017 tat ca - 1
Tính tổng
\(1^3+2^3+3^3+...+100^3\)
\(1.2^2+2.3^2+3.4^2+...+99.100^2\)
CMR:1/2+1/3.4+1/5.6+...+1/2015.2106=1/1009+1/1010+...+1/2016