1+1+2+2.....+100
(1/100-1/2^2).(1/100-1/3^2).(1/100-1/4^2)........(1/100-1/2022^2)
\(=\left(\dfrac{1}{100}-\dfrac{1}{1^2}\right)\left(\dfrac{1}{100}-\dfrac{1}{4}\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{10^2}\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{400}\right)\)
\(=\left(\dfrac{1}{100}-\dfrac{1}{100}\right)\cdot\left(\dfrac{1}{100}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{400}\right)\)
\(=0\cdot\left(\dfrac{1}{100}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{400}\right)=0\)
1+1+1+1+1+2+2+2+2+2+......+99+99+99+99+99+100+100+100+100+100=?
tu 1 den 100 co 100 so
nen tong cac so do la : ( 100 + 1 ) x 100 : 2 = 5050
nhin tong tren , ta thay moi so duoc lap lai 4 lan nen tong do la : 5050 x 4 = 20200
dap so : 20200
Tính có bao nhiêu số hạng: (100-1):1+1 x 5= 500(số)
Tính tổng của dãy số trên: (100+10) x 500 :2 x 5 =137500
tổng = 5050 vì mỗi số xh 4 lần nên tg = 5050*4=20200
(1/100-12).(1/100-1/22).(1/100-1/32)....(1/100-1/202)
A = (\(\dfrac{1}{100}\) - 12).(\(\dfrac{1}{100}\) - \(\dfrac{1}{2^2}\)).(\(\dfrac{1}{100}\) - \(\dfrac{1}{3^2}\))...(\(\dfrac{1}{100}\) - \(\dfrac{1}{20^2}\))
A = (\(\dfrac{1}{10^2}\) - 12).(\(\dfrac{1}{10^2}\) - \(\dfrac{1}{2^2}\)).(\(\dfrac{1}{10^2}\) - \(\dfrac{1}{3^2}\))..(\(\dfrac{1}{10^2}\) - \(\dfrac{1}{10^2}\))....(\(\dfrac{1}{10^2}\) - \(\dfrac{1}{20^2}\))
A = (\(\dfrac{1}{10^2}\) - 12).(\(\dfrac{1}{10^2}\) - \(\dfrac{1}{2^2}\)).(\(\dfrac{1}{10^2}\) - \(\dfrac{1}{3^2}\))...0.(\(\dfrac{1}{10^2}\) - \(\dfrac{1}{20^2}\))
A = 0
[1/100-1^2]ư.[1/100-(1/2)^2].[1/100-(1/3)^2]...[1/100-(1/20)^2]
Giải đầy đủ cho mik nha
CMR:
a)1/10^2 +1/11^2+1/12^2+...+1/100^2 >3/4
b)1/2^2+1/3^2+1/4^2+...+1/100^2<99/100
c)1/2^2+1/3^2+1/4^2+...+1/100^2<3/4
1+1+1+1+2+2+2+2+3+3+3+3+...+99+99+99+99+100+100+100+100=?
=========================
==========================ko bt
( ;-; )
1+(1+2)+(1+2+3)+...+(1+2+3+4+...+99+100)/(1*100+2*99+...+99*2+100*1)*2013
Ta chia thành hai vế (1) và (2)
Số số hạng (1) là :
( 101 - 1 ) : 1 + 1 = 101 ( số )
Tổng (1) là :
( 101 + 1 ) x 101 : 2 = 5151
Tự tính tiếp
\(1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+99+100\right)\)
\(=\left(1+1+1+...+1\right)+\left(2+2+...+2\right)+\left(3+...+3\right)+...+\left(99+99\right)+100\)
\(=1.100+2.99+3.98+...+99.2+100.1\)
Do đó kết quả của phép tính cần tìm là:
\(\frac{1.100+2.99+...+99.2+100.1}{\left(1.100+2.99+...+99.2+100.1\right).2013}=\frac{1}{2013}\)
Tính
A=1.(100-1)+2.(100-2)+3.(100-3)+..............+99.(100-99)
B=1.(100+1)+2.(100+2)+3.(100+3)+...........+99.(100+99)
Tính [100-(1+1/2+1/3+...+1/100)]:(1/2+2/3+...+99/100)