1-2+3-4+5-6+............+45-46+47-48+100

= (1-2)+(3-4)+(5-6)+............+(45-46)+(47-48)+100

=  (-1) + (-1) + (-1)+ ...............+(-1)+(-1) +100 ( có 24 cặp -1)

=(-1)x24+100

=( -24) +100

=76

Nhớ k cho mk nha! ^-^

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

\(3m=3+3^2+...+3^{101}\)

\(\Rightarrow3m-m=3^{101}-1\)

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

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

\(\Rightarrow\)\(3M=3+3^2+3^3+...+3^{101}\)

        \(-M=1+3+3^2+...+3^{100}\)

\(\Rightarrow\)\(2M=3^{101}-1\)

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

thanks nha

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

\(\Rightarrow3A=1-\frac{2}{3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

\(4A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

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

\(\Rightarrow3B=3+1+...+\frac{3}{3^{98}}\)

\(2B=3-\frac{1}{3^{99}}\)

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

Thay B vào 4A ta có:

\(4A=\frac{3}{2}-\frac{1}{3^{99}.2}\)

\(A=\frac{3}{2.4}-\frac{1}{3^{99}.2.4}\)

\(A=\frac{3}{8}-\frac{1}{3^{99}.8}\)

Vì \(\frac{3}{8}>\frac{3}{16}\)

\(\Rightarrow\frac{3}{8}-\frac{1}{3^{99}.8}< \frac{3}{16}\)

Vậy \(A< \frac{3}{16}\)

1-2+3-4+5-6+...+99-100+101 
= (1+3+5+...+101) - (2+4+6+...+100) 
tu 1 den 101 co : (101-1):2+1=51 
1+..+101 = (1+101)x 51:2= 2601 
tu 2 den 100 co : (100-2);2+1=50 
2+...+100 = (100 +2) x 50:2=2550 
=> A= 2601-2550=51

3I = 3¹ + 3² + 3³ + 3⁴ + ... + 3¹⁰¹

3I - I = (3¹ + 3² + 3³ + 3⁴ + ... + 3¹⁰¹) - (3⁰ + 3¹ + 3² + 3³ + ... + 3¹⁰⁰)

2I = 3¹⁰¹ - 3⁰

2I = 3¹⁰¹ - 1

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

HUY:thôi cái trò chtt vớ vẩn đó đi!!!

ai tick đến 190 thì mik tick cho cả đời