A=1/3^2+1/4^2+1/5^2+1/6^2+...+1/100^2<1/2-1/3+1/3-1/4+...+1/99-1/100

=>A<1/2-1/100<1/2

a)  \(P=\frac{1+2}{1^2.2^2}+\frac{2+3}{2^2.3^2}+...+\frac{9+10}{9^2.10^2}\)

\(P=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\) ( rút gọn số mũ nhé )

\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\)

\(P=1-\frac{1}{10}=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)

Vì \(\frac{9}{10}< 1\Rightarrow P< 1\) (đpcm)

b) Chút nữa mình làm nhé ^^

b) 

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

Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)

Ta so sánh giữa A và Q.

\(\frac{1}{1.2}>\frac{1}{3};\frac{1}{2.3}>\frac{1}{3^2};\frac{1}{3.4}>\frac{1}{3^3};....;\frac{1}{100.101}>\frac{1}{3^{100}}\)

\(\Rightarrow Q< A\)

Ta lại tiếp tục so sánh A và \(\frac{1}{2}\)

Ta có :

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)

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

\(\Rightarrow A=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\Leftrightarrow A< \frac{1}{2}\)

Ta được:

\(Q< A< \frac{1}{2}\Leftrightarrow Q< \frac{1}{2}\)

b) \(Q=\frac{1}{3}\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)\)

\(Q=\frac{1}{3}\left(1+Q-\frac{1}{3^{100}}\right)\)

\(Q=\frac{1}{3}+\frac{1}{3}Q-\frac{1}{3^{101}}\)

\(\frac{2}{3}Q=\frac{1}{3}-\frac{1}{3^{101}}\)

\(\left(=\right)Q=\left(\frac{1}{3}-\frac{1}{3^{101}}\right):\frac{2}{3}=\left(\frac{1}{3}-\frac{1}{3^{101}}\right).\frac{3}{2}\)

\(Q=\frac{1}{2}-\frac{1}{3^{100}.2}< \frac{1}{2}\) (đpcm)

2A=1-1/2+1/2^2-...+1/2^98-1/2^99

=>3A=1-1/2^100

=>\(A=\dfrac{2^{100}-1}{3\cdot2^{100}}\)

Đây nha bạn:

A=5.72​+7.125​+12.197​+19.289​+28.3911​+39.401​

=7−55.7+12−77.12+19−1212.19+28−1919.28+39−2828.39+40−3939.40=5.77−5​+7.1212−7​+12.1919−12​+19.2828−19​+28.3939−28​+39.4040−39​

=15−17+17−112+112−119+119−128+128−139+139−140=51​−71​+71​−121​+121​−191​+191​−281​+281​−391​+391​−401​

=15−140=740=

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2};\dfrac{1}{3^2}< \dfrac{1}{2\cdot3};...;\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\\ \Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}< 1\\ \RightarrowĐpcm\)

N=1/2+1/2²+...+1/2¹⁰

2N=1+1/2+...+1/2⁹

2N-N=1-1/2¹⁰=1-1/1024=1023/1024

Giải:

N=1/2+1/2²+1/2³+...+1/2⁹+1/2¹⁰

2N=1+1/2+1/2²+...+1/2⁸+1/2⁹

2N-N=(1+1/2+1/2²+...+1/2⁸+1/2⁹)-(1/2+1/2²+1/2³+...+1/2⁹+1/2¹⁰)

N=1-1/2¹⁰=1023/1024

Chúc bạn học tốt!

1.

\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}\)

\(=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+...+\dfrac{100-1}{100!}\)

\(=\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{3!}-\dfrac{1}{4!}+...+\)\(\dfrac{1}{99!}-\dfrac{1}{100!}\)

\(=1-\dfrac{1}{100!}< 1\)

2.

\(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+...+\)\(\dfrac{1}{100!}\)

Ta có:

\(=\dfrac{1.2}{2!}-\dfrac{1}{2!}+\dfrac{2.3}{3!}-\dfrac{1}{3!}+\dfrac{3.4}{4!}-\dfrac{1}{4!}+...+\)\(\dfrac{99.100}{100!}-\dfrac{1}{100}\)

\(=\left(\dfrac{1.2}{2!}+\dfrac{2.3}{3!}+\dfrac{3.4}{4!}+...+\dfrac{99.100}{100!}\right)\)\(-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}\right)\)

\(=\left(1+1+\dfrac{1}{2!}+...+\dfrac{1}{98!}\right)\)\(-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}\right)\)

\(=2-\dfrac{1}{99!}-\dfrac{1}{100!}< 2\)