Tính :
\(S=2+2^2+2^3+...+2^{100}\)
\(P=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
Những câu hỏi liên quan
Tính tổng S =\(\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
Tính S=\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{100}\left(1+2+3+...+100\right)\)
Tính S ?
Tính :
\(S=2+2^2+2^3+...+2^{100}\)
\(P=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
a)S=2+22+23+...+2100
2S=2(2+22+23+...+2100)
2S=22+23+...+2101
2S-S=(22+23+...+2101)-(2+22+23+...+2100)
S=2101-2
b)\(P=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(3P=3\left(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{100}}\right)\)
\(3P=1+\frac{1}{3}+...+\frac{1}{3^{99}}\)
\(3P-P=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)\)
\(2P=1-\frac{1}{3^{100}}\)
\(P=\left(1-\frac{1}{3^{100}}\right):2\)
Đúng 0
Bình luận (0)
S=2+22+23+...+2100
=>2S=22+23+24+...+2101
=>2S-S=(22+23+24+...+2101)-(2+22+23+...+2100)
=>S=2101-2=2(2100-1)
\(F=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(\Rightarrow3F=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(\Rightarrow3F-F=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow2F=1-\frac{1}{3^{100}}\)
\(\Rightarrow F=\frac{1-\frac{1}{3^{100}}}{2}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Tính nhanh, tính bằng cách hợp lí:
S=\(3+\frac{3}{1+2}+\frac{3}{1+2+3}+\frac{3}{1+2+3+4}+....+\frac{3}{1+2+3+...+100}\)
S=\(3\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\right)\)
\(S=3\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{5050}\right)\)
\(S=3.\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{10100}\right)\)
\(S=\frac{3}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\right)\)
\(S=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(S=\frac{3}{2}\left(1-\frac{1}{101}\right)\)
\(S=\frac{3}{2}.\frac{100}{101}=\frac{150}{101}\)
Đúng 0
Bình luận (0)
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
Chứng minh :
a) frac{1}{3}-frac{2}{3^2}+frac{3}{3^3}-frac{4}{3^4}+...+frac{99}{3^{99}}-frac{100}{3^{100}} frac{3}{16} frac{1}{3}-frac{2}{3^2}+frac{3}{3^3}+frac{4}{4^4}+...+frac{99}{3^{99}}-frac{100}{3^{100}} frac{3}{16}
b)frac{1}{41}+frac{1}{42}+frac{1}{43}+...+frac{1}{79}+frac{1}{80} frac{7}{12}
c) Cho Sfrac{3}{10}+frac{3}{11}+frac{3}{12}+frac{3}{13}+frac{3}{14}
Chứng minh 1 S 2
Đọc tiếp
Chứng minh :
a) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{4^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
b)\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{79}+\frac{1}{80}< \frac{7}{12}\)
c) Cho \(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
Chứng minh \(1< S< 2\)
tính nhanh
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
tính nhanh: \(\frac{100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}\)
Tách 100 thành 100 số 1
Ta có: TS=\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=100-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{100}=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
=\(0+\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}=\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}\)=MS
=> Phân số trên=1
Đúng 0
Bình luận (0)
Tính giá trị của biểu thức:
\(s=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+....+\frac{1}{100}\left(1+2+3+...+100\right)\)
ta được S={ }
\(S=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{100}\left(1+2+3+....+100\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+.....+\frac{1}{100}.\frac{100.101}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+.....+\frac{101}{2}\)
\(=\frac{2+3+4+....+101}{2}\)
\(=\frac{\frac{101.102}{2}-1}{2}\)
\(=2575\)
Vậy \(S=2575\)
Đúng 0
Bình luận (0)