\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)

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

     \(=1-\frac{1}{100}=\frac{99}{100}\)

1. 1 + ( -2) +3 +(-4) + .........+ 19 + (-20)

= -1 + ( -1) +....+(-1)

= -1. 10

= -10

2. 1 – 2 + 3 – 4 + . . . + 99 – 100 

= ( -1) + (-1) +....+(-1)

= -1. 50

= -50

3. 2 – 4 + 6 – 8 + . . . + 48 – 50 

= (-2) + (-2) +....+ (-2)

= -2. 12 + 26

= -24 + 26

= 2

4. – 1 + 3 – 5 + 7 - . . . . + 97 – 99

= 2 + 2 +......+2

= 2.25

= 50

5. 1 + 2 – 3 – 4 + ... + 97 + 98 – 99 - 100

= (1+2-3-4) +......+ ( 97+98-99 -100)

= -4 . (-4).....(-4)

= -4. 25

= -100

tại sao câu b lại x50

\(T=(\frac{1}{2}-\frac{1}{3})(\frac{1}{2}-\frac{1}{5})(\frac{1}{2}-\frac{1}{7}).....(\frac{1}{2}-\frac{1}{99})\)

\(\implies T=\frac{1}{2}(1-\frac{2}{3}).\frac{1}{2}(1-\frac{2}{5}).\frac{1}{2}(1-\frac{2}{7}).....\frac{1}{2}(1-\frac{2}{99})\)

Thấy T có: (99-3):2+1=49(SH)

\(\implies T=(\frac{1}{2}.49).[(1-\frac{2}{3}).(1-\frac{2}{5})...(1-\frac{2}{99})\)

\(\implies T=\frac{49}{2}.\frac{1}{99}=\frac{49}{198}\)

\(S=1+\dfrac{1}{2}+\dfrac{1}{2^3}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{101}}\)

\(\Rightarrow S-1=\dfrac{1}{2}+\dfrac{1}{2^3}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{101}}\)

\(\Rightarrow\dfrac{1}{4}\left(S-1\right)=\dfrac{1}{2^3}+\dfrac{1}{2^5}+\dfrac{1}{2^7}+...+\dfrac{1}{2^{103}}\)

\(\Rightarrow\dfrac{1}{4}\left(S-1\right)-\left(S-1\right)=\dfrac{1}{2^3}+\dfrac{1}{2^5}+\dfrac{1}{2^7}+...+\dfrac{1}{2^{103}}-\dfrac{1}{2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{101}}\)

\(\Rightarrow\dfrac{3}{4}\left(S-1\right)=\dfrac{1}{2^{103}}\)

\(\Rightarrow S-1=\dfrac{1}{2^{103}}:\dfrac{3}{4}\)

\(\Rightarrow S-1=\dfrac{4}{3.2^{103}}\)

\(\Rightarrow S=\dfrac{4}{3.2^{103}}+1\)

⇒S−1=12+123+125+...+12101⇒S−1=12+123+125+...+12101

⇒14(S−1)−(S−1)=123+125+127+...+12103−12−123−...−12101⇒14(S−1)−(S−1)=123+125+127+...+12103−12−123−...−12101

⇒S−1=12103:34⇒S−1=12103:34

⇒S=43.2103+1

dễ 

\(\infty\)

Soái ca 2k6 Làm đi bạn !!

\(\frac{3^{2}+1}{3^{2}-1}+\frac{5^{2}+1}{5^{2}-1}+...+\frac{99^{2}+1}{99^{2}-1}=49+\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}=49+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}=49.49\)

\(\frac{3^2+1}{3^2-1}+\frac{5^2+1}{5^2-1}+\frac{7^2+1}{7^2-1}+...+\frac{99^2+1}{99^2-1}\)

\(=\frac{3^2-1+2}{3^2-1}+\frac{5^2-1+2}{5^2-1}+\frac{7^2-1+2}{7^2-1}+...+\frac{99^2-1+2}{99^2-1}\)

\(=1+\frac{2}{3^2-1}+1+\frac{2}{5^2-1}+1+\frac{2}{7^2-1}+...+1+\frac{2}{99^2-1}\)

\(=\left(1+1+...+1\right)+\left(\frac{2}{\left(3-1\right)\left(3+1\right)}+\frac{2}{\left(5-1\right)\left(5+1\right)}+...+\frac{2}{\left(99-1\right)\left(99+1\right)}\right)\)

\(=49+\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{98.100}\right)\)

\(=49+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(=49+\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=49+\frac{49}{100}\)

\(=\frac{4949}{100}\)

a)\(1-2+3-4+5-6+7-8+8-9+9-10\)

=\(\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+\left(7-8\right)+\left(8-9\right)+\left(9-10\right)\)

\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)\)

\(=\left(-1\right).6\)

\(=-6\)

b)\(1-2+3-4+...+99-100\)

\(=\left(1-2\right)+\left(3-4\right)+...+\left(99-100\right)\)}\(\left[\left(100-1\right):1+1\right]:2=50\)(cặp)

\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)} 50 số (-1)

\(=\left(-1\right).50\)

\(=-50\)

c)\(1-3+5-7+9-11+13-15\)

\(=\left(1-3\right)+\left(5-7\right)+\left(9-11\right)+\left(13-15\right)\)

\(=\left(-2\right)+\left(-2\right)+\left(-2\right)+\left(-2\right)\)

\(=\left(-2\right).4\)

\(=-8\)

d)\(1-3+5-7+...-99+101\) (Đối với bài này, có vẻ đề sai, mình đã sửa lại rồi

\(=\left(1-3\right)+\left(5-7\right)+...+\left(97-99\right)+101\) } \(\left[\left(99-1\right):2+1\right]:2=25\)(cặp)

\(=\left(-2\right)+\left(-2\right)+\left(-2\right)+...+\left(-2\right)\) } 25 số (-2)

\(=\left(-2\right).25\)

\(=-50\)

e)\(-1-2-3-4-...-99-100\)

\(=\left(-1\right)+\left(-2\right)+\left(-3\right)+...+\left(-99\right)+\left(-100\right)\)

\(=\left[\left(-1\right)+\left(-100\right)\right]+\left[\left(-2\right)+\left(-99\right)\right]+...+\left[\left(-51\right)+\left(-50\right)\right]\) } \(\left[\left(100-1\right):1+1\right]:2=50\)(cặp) (phần này của đề bài, không thay được như (-100) hoặc (-1))

\(=\left(-100\right)+\left(-100\right)+\left(-100\right)+...+\left(-100\right)\)} 50 số (-100)

\(=\left(-100\right).50\)

\(=-5000\)

a, -5

b, -50

c, -8

d, -50

e, -5050

cách làm nữa nha