A = 1 + \(\frac{1}{2}\left(1+2\right)\)+ \(\frac{1}{3}\left(1+2+3\right)\)+ .... + \(\frac{1}{100}\left(1+2+3+...+100\right)\)

A = \(1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+...+\frac{1}{100}\cdot\frac{100.101}{2}\)

A = \(\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{101}{2}\)

A = \(\frac{2+3+4+...+101}{2}\)

A = \(\frac{\left(101+2\right).100}{2}\div2\)

A  = \(5150\div2=2575\)

Tinh nhanh

(100-1²).(100-2²).(100-3²)...(100-19²)

A=1+3/2^3+4/2^4+5/2^5+...100/2^100
1/2*A = 1/2 + 3/2^4 + 4/2^5 +....+ 99/2^100 + 100/2^101

A- A/2 = 1/2A =1/2 + 3/2^3 + 1/2^4 +...+1/2^100 - 100/2^101=

= [1/2+1/2^2 +1/2^3 +...+1/2^100] -100/2^101 (Do 3/2^3 = 1/2^2 +1/2^3)

=[1-(1/2)^101]/(1-1/2) -100/2^101 =

=(2^101 -1)/2^100 - 100/2^101

=> A= (2^101 -1)/2^99 - 100/2^100

minh biet lam ne nhung ban phai cho minh nhe

ai giup minh lam bai nay voi 

thanks nhieu

\(A=\frac{1}{1}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)

\(\frac{1}{2}A=\frac{1}{2}.\left(\frac{1}{1}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\right)\)

\(\frac{1}{2}A=\frac{1}{2}+\frac{3}{2^4}+\frac{4}{2^5}+...+\frac{100}{2^{101}}\)

\(\frac{1}{2}A-A=\left(1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\right)-\left(\frac{1}{2}+\frac{3}{2^4}+\frac{4}{2^5}+...+\frac{100}{2^{101}}\right)\)

\(\frac{1}{2}A=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}-\frac{1}{2}-\frac{3}{2^4}-\frac{4}{2^5}-...-\frac{100}{2^{101}}\)

\(\frac{1}{2}A=\frac{1}{2}+\frac{3}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)

\(\frac{1}{2}A=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^{100}}\right)-\frac{100}{2^{101}}\)

\(\frac{1}{2}A=\frac{\left[\frac{1}{2}-\left(\frac{1}{2}\right)^{101}\right]}{\frac{1}{2}}-\frac{100}{2^{101}}\)

A=2

Bài 1 :

\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)

Ta thấy :

\(3=2^2-1\)

\(15=4^2-1\)

\(35=6^2-1\)

.....

\(9999=100^2-1\)

\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)

\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)

\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)

nhanh len nhé mik đang cần gấp ai lam trước mik tích cho

Bài 6 :

\(C=1^2+2^2+...+100^2=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}=\dfrac{100.101.201}{6}=338350\)

Bài 9 :

\(S=1^2+2^2+3^2+...+99^2=\dfrac{99.\left(99+1\right)\left(2.99+1\right)}{6}=\dfrac{99.100.199}{6}=328350\)

yêu cầu bạn ơi?

\(G=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)

\(3G=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)

\(3G-G=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)\(-\frac{1}{3}-\frac{2}{3^2}-\frac{3}{3^3}-...-\frac{100}{3^{100}}\)

\(2G=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt \(M=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

\(3M=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}\)

\(3M-M=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}\)\(-1-\frac{1}{3}-\frac{1}{3^2}-...-\frac{1}{3^{99}}\)

\(2M=3-\frac{1}{3^{99}}\Leftrightarrow M=\frac{3}{2}-\frac{1}{3^{99}.2}\)

\(\Rightarrow2G=\frac{3}{2}-\frac{1}{3^{99}.2}-\frac{100}{3^{100}}\)

\(\Rightarrow G=\frac{3}{4}-\frac{1}{3^{99}.2^2}-\frac{100}{3^{100}.2}\)