Question

tính tổng

A= \(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{197\times199}\)

B= \(1+2+4+8+16+...+512+1024\)

Answer 1

a, \(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{197.199}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{197}-\dfrac{1}{199}\)

\(=\dfrac{1}{3}-\dfrac{1}{199}=\dfrac{196}{597}\)

b, \(B=1+2+4+...+1024\)

\(\Rightarrow2B=2+4+8+...+2056\)

\(\Rightarrow2B-B=\left(2+4+8+...+2056\right)-\left(1+2+4+...+1024\right)\)

\(\Rightarrow B=2056-1=2055\)

Answer 2

\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{197}-\dfrac{1}{199}\)

\(A=\dfrac{1}{3}-\dfrac{1}{199}\)

\(A=\dfrac{199}{597}-\dfrac{3}{597}=\dfrac{196}{597}\)

Answer 3

\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{197.199}\)

\(A=\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+...+\dfrac{199-197}{197.199}\)

\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{197}-\dfrac{1}{199}\)

\(A=\dfrac{1}{3}-\dfrac{1}{999}\)

\(A=\dfrac{196}{697}\)

\(B=1+2+4+8+16+...+512+1024\)

\(2B=2+4+8+32+...+1024+2048\)

\(B=\left(2+4+8+...+2048\right)-\left(1+2+4+...+1024\right)\)

\(B=2048-1\)

\(B=2047\)

Answer 4

\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{197.199}\)

\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{197}-\dfrac{1}{199}\)

\(A=\dfrac{1}{3}-\dfrac{1}{199}=\dfrac{196}{597}\)

\(B=1+2+4+8+16+...+512+1024\)

\(B=2^0+2^1+2^2+2^3+2^4+...+2^9+2^{10}\)

\(2B=2\left(2^0+2^1+2^2+2^3+2^4+...+2^9+2^{10}\right)\)

\(2B=2^1+2^2+2^3+2^4+2^5+...+2^{10}+2^{11}\)

\(2B-B=\left(2^1+2^2+2^3+2^4+2^5+...+2^{10}+2^{11}\right)-\left(2^0+2^1+2^2+2^3+2^4+...+2^9+2^{10}\right)\)

\(B=2^{11}-2^0\)

\(B=2048-1=2047\)