Đại số lớp 6
Các câu hỏi tương tự
CMR 1/3-2/3^2+3/2^3-....+100/2^100<3/16
CMR:1/3-2/3^2+3/3^3-4/3^4+.......+99/3^99-100/3^100<3/16
CMR: 100- \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}\)
CMR 100-(1+\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\))= (\(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}\))
CMR \(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}< 1\)
Cmr : \(\dfrac{1}{3}\) - \(\dfrac{2}{3^2}\) +\(\dfrac{3}{3^3}\) - \(\dfrac{4}{3^4}\) + ...+\(\dfrac{99}{3^{99}}\) - \(\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
a ) A = 1/4 + 1/4^2 +1/4^3 +.........+ 1/4^100 + 1/3.4^100
b) B = 1/3 - 1/3^2 + 1/3^3 - 1/3^4 +.........+ 1/ 3^99
Cho S=\(1-3+3^2-3^3+...+3^{98}-3^{99}\)
a, CMR: S là bội của -20
b,Tính S từ đó suy ra \(3^{100}\) chia cho 4 dư 1
1. 1+ (-2) + 3+ (-4) + . . . +19 + (-20)
2. 1 - 2 + 3- 4 + . . . + 99 - 100
3. -1 + 3 -5 + 7 - . . . +97 - 99
4. 1+ 2 - 3+ 4 + . . . +97 + 98 - 99 - 100