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  = \(5150\div2=2575\)

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Câu hỏi của ❖︵Ňɠυүễη Çɦâυ Ƭυấη Ƙїệт♔ - Toán lớp 7 - Học toán với OnlineMath

=100/3^2+2+3+4

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