1+2+2^2+............+2^1001
Tính tổng *cách làm *
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Hiệu số phần bằng nhau là :
4 - 1 = 3 ( phần )
Số bé là :
252 : 3 = 84
Số lớn là :
252 + 84 = 336
Đáp số : Số bé : 84
Số lớn : 336
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Đặt \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+....+\dfrac{1}{\left(n-2\right)\cdot\left(n-1\right)\cdot n}\)
\(2A=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{\left(n-2\right)\cdot\left(n-1\right)\cdot n}\\ 2A=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+....+\dfrac{1}{\left(n-2\right)\cdot\left(n-1\right)}-\dfrac{1}{\left(n-1\right)\cdot n}\)
\(2A=\dfrac{1}{1\cdot2}-\dfrac{1}{\left(n-1\right)\cdot\left(n-2\right)}\)
\(A=\dfrac{1}{4}-\dfrac{1}{\left(n-1\right)\cdot\left(n-2\right)\cdot2}\)
\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\cdot\cdot\cdot+\dfrac{1}{\left(n-2\right)\cdot\left(n-1\right)\cdot n}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\cdot\cdot\cdot+\dfrac{2}{\left(n-2\right)\cdot\left(n-1\right)\cdot n}\right)\)
\(=\dfrac{1}{2}\left[\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+\cdot\cdot\cdot+\dfrac{1}{\left(n-2\right)\left(n-1\right)}-\dfrac{1}{\left(n-1\right)n}\right]\)
\(=\dfrac{1}{2}\left[\dfrac{1}{1\cdot2}-\dfrac{1}{\left(n-1\right)n}\right]\)
\(=\dfrac{1}{2}\cdot\left[\dfrac{n\left(n-1\right)}{2n\left(n-1\right)}-\dfrac{2}{2n\left(n-1\right)}\right]\)
\(=\dfrac{1}{2}\cdot\dfrac{n\left(n-1\right)-2}{2n\left(n-1\right)}\)
\(=\dfrac{n^2-n-2}{4n\left(n-1\right)}\)
#\(Toru\)
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}\)
\(=\frac{6}{7}\)
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}\)
\(=\frac{6}{7}\)
Ta có:
\(1+2+4+8+...+512=1+2+2^2+2^3+...+2^9\)
Đặt \(A=1+2+2^2+2^3+...+2^9\)
\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^9+2^{10}\)
\(\Rightarrow2A-A=2^{10}-1\)
\(\Rightarrow A=2^{10}-1=1024-1=1023\)