\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{47}-\frac{1}{48}+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+....+\frac{1}{25}\right)\)\(=\frac{1}{26}+...+\frac{1}{50}\)

1+12=13

giup mih voi

So sánh tổng : S = 1/5 + 1/9 + 1/10 + 1/41 + 1/42 với 1/2

S=

=50/50+50/49+50/48+...+50/2

=50.(1/50+1/49+1/48+...+1/4+1/3+1/2)

=50

P=

P=(1/49+1)+(2/48+1)+...+(48/2+1)+1

P= 50/49+50/48+....+50/2+50/50=1

vậy s/p = 1/50

A = - ( 5 - 6 ) - ( 3 - 4 + 5 - 7 )

A = -5 + 6 - 3 + 4 - 5 + 7

A = ( 6 + 4 ) + ( -5 + (-5) ) + ( -3 + 7 )

A = 10 + (-10) + 4

A = 0 + 4

A = 4

P = ( 1 + 3 + 5 + ... + 47 + 49 ) - ( 2 + 4 + 6 + ... + 48 + 50 )

P = \(\frac{\left(1+49\right)\cdot\left(\left(49-1\right):2+1\right)}{2}\)  -  \(\frac{\left(2+50\right)\cdot\left(\left(50-2\right):2+1\right)}{2}\)

P = \(625-650\)

P = \(-25\)

=> \(A=\frac{\left(\frac{49}{1}+\frac{48}{2}+...+\frac{1}{49}\right)}{50}=\frac{49}{50.1}+\frac{48}{50.2}+...+\frac{1}{50.49}\)

=> \(A=\frac{50-1}{50.1}+\frac{50-2}{50.2}+...+\frac{50-49}{50.49}\)

=> \(A=\left(\frac{50}{50.1}+\frac{50}{50.2}+...+\frac{50}{50.49}\right)-\left(\frac{1}{50.1}+\frac{2}{50.2}+...+\frac{49}{50.49}\right)\)

=> \(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\) ( có 49 số 1/50 )

=> \(A=1+\frac{1}{2}+...+\frac{1}{49}-\frac{49}{50}=\left(1-\frac{49}{50}\right)+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\)

=> \(A=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\)

Vậy A không phải là số tự nhiên 

Số số hạng của dãy :

( 99 - 1 ) : 2 + 1 = 50 số

Mỗi cặp có 2 số hạng

⇒ Có số cặp là : 50 : 2 = 25 cặp

Mỗi cặp có kết quả = 2

⇒ Kết quả = 2 . 25 = 50