Ta có: \(S=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>\(3S=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\ldots+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

=>3S+S=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{99}{3^{98}}-\frac{100}{3^{99}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>4S=\(1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt \(A=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>3A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}\)

=>3A+A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>4A=\(-1-\frac{1}{3^{99}}=\frac{-3^{99}-1}{3^{99}}\)

=>\(A=\frac{-3^{99}-1}{4\cdot3^{99}}\)

Ta có: \(4S=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(=1+\frac{-3^{99}-1}{4\cdot3^{99}}-\frac{100}{3^{100}}=1+\frac{-3^{100}-3-400}{4\cdot3^{100}}=1-\frac14-\frac{403}{4\cdot3^{100}}<\frac34\)

=>\(S<\frac{3}{16}\)

mà 3/16<3/15=1/5

nên S<1/5

\(32^{15}=\left(2^5\right)^{15}=2^{5.15}=2^{75}\)

\(4^{39}=\left(2^2\right)^{39}=2^{2.39}=2^{78}\)

Do \(2^{78}>2^{75}\)

\(\Rightarrow4^{39}>32^{15}\)

\(\Rightarrow1+4+4^2+...+4^{39}>32^{15}\)

\(\Rightarrow3\left(1+4+4^2+...+4^{39}\right)>32^{15}\)

Vậy \(A>B\)

mọi ng giúp mik với

1/

6 = 1*2*3

24 = 2*3*4

.......

Số thứ 100: 100*101*102 

TỔng dãy trên là A thì bằng:

A = 1*2*3 + 2*3*4 + ..... + 100*101*102

4A = 1*2*3*4 + 2*3*4*4 + .... + 100*101*102*4

4A = [1*2*3*4 - 0*1*2*3]+ [2*3*4*5 - 1*2*3*4]+ ...+[100*101*102*103 - 99*100*101*102]

4A = 0*1*2*3 + [1*2*3*4-1*2*3*4]+[2*3*4*5-2*3*4*5]+..........+[99*100*101*101-99*100*101*102] + 100*101*102*103

4A = 100*101*102*103

A = 25*101*102*103 = 26527650

2/

\(A=\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{73\cdot76}=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{73}-\frac{1}{76}\)

\(A=\frac{1}{4}-\frac{1}{76}=\frac{9}{38}\)

P/s: Vì tử bằng khoẳng cách dưới mẫu nên ta có thể rút gọn nhanh như vậy

2)    \(A=\frac{3}{4.7}+\frac{3}{7.10}+........+\frac{3}{73.76}\)

       \(A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.......+\frac{1}{73}-\frac{1}{76}\)

       \(A=\frac{1}{3}-\frac{1}{76}\)

      \(A=\frac{73}{228}\)

\(\frac{1}{2.2}< \frac{1}{1.2}\)

\(\frac{1}{3.3}< \frac{1}{2.3}\)

......

\(\frac{1}{100.100}< \frac{1}{99.100}\)

\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{100.100}< \frac{1}{1.2}+..+\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{2.2}+..+\frac{1}{100.100}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow\frac{1}{2.2}+..+\frac{1}{100.100}< 1-\frac{1}{100}< 1\).Suy ra điều phải chứng minh. câu b tương tự. bấm đúng cho mình nha

B<3\4 là đúng

khó thế

\(A=4+2^2+2^3+...+2^{99}\)

=>  \(2A=8+2^3+2^4+...+2^{100}\)

=>  \(2A-A=\left(8+2^3+2^4+...+2^{100}\right)-\left(4+2^2+2^3+...+2^{99}\right)\)

=>  \(A=2^{100}< 2^{200}=2^{2.100}=4^{100}=B\)

Vậy  A < B

mình chỉ làm đc câu a và d thôi bạn có **** k? nếu **** thì liên hệ mình làm cho