1/ $\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}$
Tính giá trị biểu thức$\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}$
Started By whitemiss, 14-08-2013 - 07:42
#2
Posted 14-08-2013 - 07:43
Zaraki
-
- Phó Quản lý Toán Cao cấp
-
- 4273 posts
PQT
Chú ý hai công thức $1+2+ \cdots + n= \frac{n(n+1)}{2}$ và $\frac{1}{n(n+1)}= \frac 1n - \frac{1}{n+1}$.
- letankhang likes this
Discovery is a child’s privilege. I mean the small child, the child who is not afraid to be wrong, to look silly, to not be serious, and to act differently from everyone else. He is also not afraid that the things he is interested in are in bad taste or turn out to be different from his expectations, from what they should be, or rather he is not afraid of what they actually are. He ignores the silent and flawless consensus that is part of the air we breathe – the consensus of all the people who are, or are reputed to be, reasonable.
Grothendieck, Récoltes et Semailles (“Crops and Seeds”).
#3
Posted 14-08-2013 - 08:07
letankhang
-
- Thành viên
-
- 1079 posts
$\sqrt{MF}'s$ $member$
1/ $\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}$
Ta xét dạng tổng quát :
$\frac{1}{1+2+3+...+n}=\frac{1}{\frac{n(n+1)}{2}}=\frac{2}{n(n+1)}=2(\frac{1}{n}-\frac{1}{n+1})$
Áp dụng vài biểu thức trên :
$1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}=2(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100})=2(1-\frac{1}{100})=\frac{99}{50}$
- Yagami Raito, Tienanh tx and whitemiss like this
$\mathfrak Lê $ $\mathfrak Tấn $ $\mathfrak Khang $ $\mathfrak tự$ $\mathfrak hào $ $\mathfrak là $ $\mathfrak thành $ $\mathfrak viên $ $\mathfrak VMF $
$\textbf{Khi đọc một quyển sách; tôi chỉ ráng tìm cái hay của nó chứ không phải cái dở của nó.}$
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users
