\(\frac{N}{2}=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)

\(\frac{N}{2}=N-\frac{N}{2}=\frac{1}{2}-\frac{1}{2^{100}}\Rightarrow N=1-\frac{1}{2^{99}}

vì 1/2+1/2 =1/2 bình nên A<1

A=[1/1+1/2+....+1/98]*2*4*...*98*3*33=A=[1/1+1/2+....+1/98]*2*4*....*98*99\(⋮\)99

\(A=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right)\times2\times3\times4\times...\times98\)

\(A=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right)\times2\times3\times4\times...\times33\times...\times98\)

\(A=\left(3\times33\right)\times\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right)\times2\times4\times...\times98\)

\(A=99\times\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right)\times2\times4\times...\times98\)

Vậy \(A⋮99\)(Vì A có thừa số 99)

đợi mình tí bài này minh làm rồi

Ta có\(M=\left[\left(1+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{50}\right)\right].2.3...98\)

\(=\left[\frac{99}{1.98}+\frac{99}{2.97}+...+\frac{99}{49.50}\right].2.3...98=99\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right).2.3...98\)

\(=99\left(\frac{k_1+k_2+...+k_{49}}{1.2.3...98}\right).2.3...98\left(k_1,k_2...k_{49}\varepsilonℕ^∗\right)=99\left(k_1+k_2+...+k_{49}\right)⋮99\Rightarrow M⋮99\left(đpcm\right)\)

B=1/2 +(1/2 )^2+(1/3 )^3+......+(1/2 )\(^{99}\)

⇒2B=1+1/2 +1/22 +......+1/298 

⇒B=2B−B=1−1/2\(^{99}\)

⇒1−1/2\(^{99}\) <1⇒B<1

\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)

=> \(2B-B=\left(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{98}\right)\)\(-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)\)

=> \(B=1-\frac{1}{2^{99}}< 1\)

\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^3+......+\left(\frac{1}{2}\right)^{99}\)

\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{98}}\)

\(\Rightarrow B=2B-B=1-\frac{1}{2^{99}}\)

\(\Rightarrow1-\frac{1}{2^{99}}< 1\Rightarrow B< 1\)

Nhanh nhanh nha

nhanh nhanh nha

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