Question

[Ben Drowned]

tính 

A=\(1^2+2^2+3^2+...+n^2\)

B= \(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{19}{2^{19}}\)

Answer 1

[alibaba nguyễn]

\(A=1^2+2^2+...+n^2\)

\(=1\left(2-1\right)+2\left(3-1\right)+...+n\left(n+1-1\right)\)

\(=\left(1.2+2.3+...+n\left(n+1\right)\right)-\left(1+2+...+n\right)\)

\(=\frac{\left[1.2.\left(3-0\right)+2.3.\left(4-1\right)+...+n.\left(n+1\right)\left(\left(n+2\right)-\left(n-1\right)\right)\right]}{3}-\left(1+2+....+n\right)\)

\(=\frac{1.2.3-0.1.2+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)}{3}-\frac{n\left(n+1\right)}{2}\)

\(=\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}\)

\(=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

Answer 2

[alibaba nguyễn]

\(2B=1+1+\frac{3}{2^2}+...+\frac{19}{2^{18}}\)

\(=2+\left(\frac{1}{2^2}+\frac{2}{2^2}+\frac{1}{2^3}+\frac{3}{2^3}+...+\frac{1}{2^{18}}+\frac{18}{2^{18}}\right)\)

\(=2+\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{18}}\right)+\left(\frac{2}{2^2}+\frac{3}{3^3}+...+\frac{18}{2^{18}}\right)\)

\(=\frac{3}{2}-\frac{19}{2^{19}}+\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{18}}\right)+\left(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{3^3}+...+\frac{18}{2^{18}}+\frac{19}{2^{19}}\right)\)

\(\Rightarrow2B=B+\frac{3}{2}-\frac{19}{2^{19}}+\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{18}}\right)\)

\(\Rightarrow B=\frac{3}{2}-\frac{19}{2^{19}}+\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{18}}\right)\left(1\right)\)

Giờ tính \(C=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{18}}\)

\(\Rightarrow2C=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{17}}\)

\(\Rightarrow2C-C=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{17}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{18}}\right)\)

\(\Rightarrow C=\frac{1}{2}-\frac{1}{2^{18}}\left(2\right)\)

Từ (1) và (2) 

\(\Rightarrow B=\frac{3}{2}-\frac{19}{2^{19}}+\frac{1}{2}-\frac{1}{2^{18}}\)

\(\Rightarrow B=2-\frac{19}{2^{19}}-\frac{1}{2^{18}}\)