1 tính
a, S= \(3+3^2+3^3+....+3^{100}\)
b, M= \(\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{100}}\)
Những câu hỏi liên quan
Tính:
\(S=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{100}\left(1+2+3+...+100\right)\)
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: \(A=1+\frac{3}{2^3}+\frac{4}{2^4}+....+\frac{100}{2^{100}}\)
Câu 1 :Tính tổng Sfrac{1}{2}+frac{1}{4}+frac{1}{8}+......+frac{1}{2^{100}}Câu 2 :Tính tổng Mfrac{3}{5}+frac{3}{5^2}+..........+frac{3}{5^{201}}Câu 3 :Tính tổng N10130+10131+10132+.......+101101Câu 4 :Tính tổng A23+43+63+........+20123Câu 5 :Tính tổng B13+33+53+........+20113Câu 6 :Tính tổng C2*4*6*8+4*6*8*10+.......+100*102*104*106 Giúp mik vs m.n !!
Đọc tiếp
Câu 1 :Tính tổng S=\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{8}\)+......+\(\frac{1}{2^{100}}\)
Câu 2 :Tính tổng M=\(\frac{3}{5}+\frac{3}{5^2}+..........+\frac{3}{5^{201}}\)
Câu 3 :Tính tổng N=10130+10131+10132+.......+101101
Câu 4 :Tính tổng A=23+43+63+........+20123
Câu 5 :Tính tổng B=13+33+53+........+20113
Câu 6 :Tính tổng C=2*4*6*8+4*6*8*10+.......+100*102*104*106
Giúp mik vs m.n !!
Chứng minh rằng:a. frac{1}{3^2}+frac{2}{3^3}+frac{3}{3^4}+frac{4}{3^5}+...+frac{99}{3^{100}}+frac{100}{3^{101}} frac{1}{4}b.frac{1}{2}-frac{1}{4}+frac{1}{8}-frac{1}{16}+frac{1}{32}-frac{1}{64} frac{1}{3}c.frac{1}{3}-frac{2}{3^2}+frac{3}{3^3}-frac{4}{3^4}+...+frac{99}{3^{99}}-frac{100}{3^{100}} frac{1}{16}d. frac{1}{5^2}-frac{2}{5^3}+frac{3}{5^4}-frac{4}{5^5}+...+frac{99}{5^{100}}-frac{100}{5^{101}} frac{1}{36}
Đọc tiếp
Chứng minh rằng:
a. \(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{99}{3^{100}}+\frac{100}{3^{101}}< \frac{1}{4}\)
b.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
c.\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{1}{16}\)
d. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)
\(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+\frac{3}{1+2+3+4}+...+\frac{3}{1+2+3+...+100}\)
Tính A
Chào bạn, bạn hãy theo dõi câu trả lời của mình nhé!
Theo mình thì đề phải là \(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+\frac{3}{1+2+3+4}+...+\frac{3}{1+2+3+...+100}\).
Ta có :
\(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+\frac{3}{1+2+3+4}+...+\frac{3}{1+2+3+...+100}\)
\(=>A=3\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+100}\right)\)
Đặt \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+100}\) là B. Ta có :
\(B=\)\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+100}\)
\(=>B=\frac{1}{1}+\frac{1}{\left(1+2\right)\cdot2:2}+\frac{1}{\left(1+3\right)\cdot3:2}+\frac{1}{\left(1+4\right)\cdot4:2}+...+\frac{1}{\left(1+100\right)\cdot100:2}\)
\(=>B=\frac{1}{1}+\frac{1}{3\cdot2:2}+\frac{1}{4\cdot3:2}+\frac{1}{5\cdot4:2}+...+\frac{1}{101\cdot100:2}\)
\(=>B=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{100\cdot101}\)
\(=>B=2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{100\cdot101}\right)\)
\(=>B=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=>B=2\left(1-\frac{1}{101}\right)\)
\(=>B=2\cdot\frac{100}{101}=\frac{200}{101}\)
\(=>A=3B=3\cdot\frac{200}{101}=\frac{600}{101}\)
Chúc bạn học tốt!
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Tính nhanh:
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+100}\)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\)
\(A=\frac{1}{3}+\frac{1}{6}+...+\frac{1}{5050}\)
\(A=2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\right)\)
\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{101}\right)\)
Tự tính
Đúng 0
Bình luận (0)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{5050}\)
\(=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{10100}\right)\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=2.\frac{99}{202}\)
\(=\frac{99}{101}\)
Đúng 0
Bình luận (0)
Tính nhanh :
\(S=\frac{1}{100}-\frac{2}{100}+\frac{3}{100}-\frac{4}{100}+\frac{5}{100}-....-\frac{98}{100}+\frac{99}{100}-\frac{100}{100}\)
\(S=\frac{1}{100}-\frac{2}{100}+\frac{3}{100}-...-\frac{98}{100}+\frac{99}{100}-\frac{100}{100}\)
\(=\frac{1-2+3-...-98+99-100}{100}\)
\(=\frac{\left[\left(1-2\right)+\left(3-4\right)+...+\left(97-98\right)+\left(99-100\right)\right]}{100}\)
\(=\frac{-1-1-1-...-1}{100}=\frac{-1.50}{100}=\frac{-50}{100}=\frac{-1}{2}\)
Vậy S=\(\frac{-1}{2}\)
\(S=\frac{1}{100}-\frac{2}{100}+\frac{3}{100}-\frac{4}{100}+\frac{5}{100}-...-\frac{98}{100}+\frac{99}{100}\)
\(S=\frac{\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+....+\left(97-98\right)+\left(99-100\right)}{100}\)
\(S=\frac{-1+\left(-1\right)+\left(-1\right)+.....+\left(-1\right)+\left(-1\right)}{100}\)
Từ 1 đến 100 có 100 số số hạng => Có 50 cặp => có 50 số (-1)
=> \(S=\frac{50\cdot\left(-1\right)}{100}=\frac{-50}{100}=\frac{-1}{20}\)
CMR
a)A=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+....+\frac{100}{3^{100}}< \frac{3}{4}\)
b)B=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+.....+\frac{100}{4^{100}}< \frac{4}{9}\)