dễ 

\(\infty\)

Soái ca 2k6 Làm đi bạn !!

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

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

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

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

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

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

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

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

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

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

Câu 1: Làm tri tiết để hiểu nha:

Ta có  : S = 3 + 5 + 7 + … + 999

Tổng dãy số tự nhiên lẻ cách đều 2 đơn vị : S = 499*(3 +999) : 2 = 249999

Câu 2:

Ta có: :B=1+11+2+...+99=1+2+..+9+11+22+..+99=9*(1+9):2 + 9*(11+99) :2= 540

Câu 3: 

C= 3+7+11+...+99 = [(99-3)/(7-4) + 1] * (3+99) : 2 =1734

Câu 4:

Ta có D= 1-2+3-4+5-6+...+99-100+101 

= (1+3+5+...+101) - (2+4+6+...+100) 

từ 1 đến 101 có : (101-1):2+1=51 

1+..+101 = (1+101)x 51:2= 2601 

từ  2 đến 100 có : (100-2);2+1=50 

2+...+100 = (100 +2) x 50:2=2550 

=> A= 2601-2550=51

Mình đã gặp bài này rồi .mình sẽ nói cho ban,nhưng bạn phải .Đáp án là:51

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

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

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

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

=-1 + (3 - 5) + (7 - 9) +...+ (97 - 99)

=-1 + -2 + -2 + ...+ -2

=(99 - 3) : 2 + 1

=49 : 2
=24.5 . -2

=-49 + -1

=-50

b.1 + 2 - 3 - 4 + ...+ 97 + 98 - 99 - 100

=1 + [(2 - 3 - 4) + 5] + ...+[(94 - 95 - 96) + 97] + (98 - 99 - 100)

=1+ 0 + ...+ 0 + -101

=-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

Đặt \(S=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+....+\frac{1}{2^{99}}\)

\(\Rightarrow\frac{1}{2^2}S=\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+.....+\frac{1}{2^{101}}\)

\(\Rightarrow S-\frac{1}{4}S=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+....+\frac{1}{2^{99}}-\frac{1}{2^3}-\frac{1}{2^5}-\frac{1}{2^7}-....-\frac{1}{2^{101}}\)

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

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

Vậy \(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+....+\frac{1}{2^{99}}=\frac{3}{2}-\frac{3}{2^{101}}\)

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

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

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

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

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

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

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

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

⇒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