Tính tổng:
a) A= 1^2*2 + 2^2 *3 + 3^2*4 +...+ 99^2*100
b) B= 1*2^2 + 2*3^2 + 3*4^2 +...+ 99*100^2
c) C= 1^3 + 2^3 + 3^3 +...+ 99^3
1.Tính
B=1/2+2/2^2+3/2^3+4/2^4+.....+99/2^99+100/2^100
tính 1/3 - 2/3^2 + 3/3^3 - 4/3^4..... + 99/3^99 - 100/3^100
Tính tổng:
\(A=1+3+3^2+3^3+...+3^{99}+3^{100}\)100
\(B=1-2+2^2-2^3+2^4-...-2^{99}+2^{100}\)
tính nhanh (1+2+3+...+99+100).(1/2-1/3-1/7-1/9)(63.1,2-21.3,6)/1-2+3-4+...+99-100
tính 1/3 - 2/3^2 + 3/3^3 - 4/3^4..... + 99/3^99 - 100/3^100 . Giúp nhanh với ạ.
Tính \(A=\frac{1}{2.\sqrt{1}+1.\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+....+\frac{1}{100.\sqrt{99}+99.\sqrt{100}}\)
c/m
a/
1/2!+2/3!+3/4!+...+99/100!<1
b/
1*2-1/2!+2*3-1/3!+3*4-1/4!+...+99*100-1/100!<2
1/ Cho A= \(\dfrac{1}{3}\)-\(\dfrac{2}{3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+.....+\(\dfrac{99}{3^{99}}\)-\(\dfrac{100}{3^{100}}\) Chứng minh A < \(\dfrac{3}{16}\)
2/ Cho B=(\(\dfrac{1}{2^2}\)-1)(\(\dfrac{1}{3^2}\)-1)....(\(\dfrac{1}{100^2}\)-1) So sánh B và \(\dfrac{-1}{2}\)