Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A = 1*2*3 + 2*3*4 + 3*4*5 ... + 99*100*101
=> 4A = 1*2*3*4 + 2*3*4*4 + 3*4*5*4 + ... +99*100*101*4
=> 4A = 1*2*3*4 + 2*3*4*(5 - 1) + 3*4*5*( 6 - 2) + ... + 99*100*101*(102 - 98)
=> 4A = 1*2*3*4 + 2*3*4*5 - 1*2*3*4 + 3*4*5*6 - 2*3*4*5 + ... + 99*100*101*102 - 98*99*100*101
=> 4A = 99*100*101*102
=> 4A = 101989800
=> A = 25497450
M*N=1/2*3/4*5/6*..*99/100*2/3*4/5*6/7*..... = 1/101 (1)
Mặt khác :
1/2 <2/3
3/4<4/5
........
99/100 < 100/101
=>1/2*3/4*5/6*....*99/100 < 2/3*4/5*6/7*....*100/101
hay M< N =>M*M<M*N hay M^2 < 1/101 <1/100
=>M^2 < 1/100 hay M^2 < (1/10)^2 =>M<1/10 (vì M>0 ) (đpcm)
\(A=\dfrac{101\cdot\dfrac{102}{2}}{\left(101-100\right)+99-98+...+3-2+1}\)
\(=\dfrac{101\cdot51}{1+1+...+1}=\dfrac{101\cdot51}{51}=101\)
\(B=\dfrac{37\cdot43\left(101-101\right)}{2+4+...+100}=0\)
a, \(A=\dfrac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)
Ta có: \(T=101+100+99+98+...+3+2+1\) \(=\dfrac{\left(101+1\right).101}{2}\)
\(=\dfrac{102.101}{2}\Leftrightarrow51.101\)
\(M=101-100+99-98+...+3-2+1\)
Ta có: \(101:2=50\) (dư \(1\))
\(\Rightarrow M=\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1\)
Có \(50\) dấu ngoặc tròn "\(\left(\right)\)"
\(\Rightarrow M=1+1+...+1+1=51.1=51\)
\(M\) có \(51\) số \(1\)
\(\Rightarrow A=\dfrac{T}{M}=\dfrac{51.101}{51}=101\)
Vậy \(A=101\)
b, \(B=\dfrac{3737.43-4343.37}{2+4+6+...100}\)
Ta có: \(T=3737.43-4343.37\)
\(T=37.101.43-43.101.37\)
\(T=0\)
\(\Rightarrow\) \(B=\dfrac{T}{2+4+6+...+100}=\dfrac{0}{2+4+6+...+100}\) \(=0\)
Vậy \(B=0\)
1 - 2 - 3 + 4 + 5 - 6 - 7 + 8+ ... + 1993 - 1994
= ( 1 - 2 - 3 + 4 ) = ( 5 - 6 - 7 + 8 ) + ... + 1993 - 1994
= 0 + 0 + ... + 1993 - 1994
= 0 + ( -1 ) = -1
b) ta có 1^2+2^2+...+n^2 = n(n+1)(2n+1)/6
=>2^2+4^2+...+(2n)^2= 2^2(1^2+2^2+...+n^2)= 2n(n+1)(2n+1)/3
và 1^2+2^2+...+(2n+1)^2=(2n+1)(2n+2)(4n+3)/...
=>1^2+3^2+5^2+...+(2n+1)^2 = (2n+1)(2n+2)(4n+3)/6 - 2n(n+1)(2n+1)/3 = (2n+1)(n+1)(2n+3)/3
=>1^2-2^2+3^2-4^2+..... -(2n)^2+(2n+1)^2 = (2n+1)(n+1)(2n+3)/3 - 2n(n+1)(2n+1)/3 = (n+1)(2n+1)
do đó ta có khi n = 100 thì
1^2-2^2+3^2-4^2.....+99^2-100^2+101^2 = (100+1)*(2*100+1)=201*101
Mình cũng không chắc câu b cho lắm
ta có 1^2+2^2+...+n^2 = n(n+1)(2n+1)/6
=>2^2+4^2+...+(2n)^2= 2^2(1^2+2^2+...+n^2)= 2n(n+1)(2n+1)/3
và 1^2+2^2+...+(2n+1)^2=(2n+1)(2n+2)(4n+3)/6
=>1^2+3^2+5^2+...+(2n+1)^2 = (2n+1)(2n+2)(4n+3)/6 - 2n(n+1)(2n+1)/3 = (2n+1)(n+1)(2n+3)/3
=>1^2-2^2+3^2-4^2+..... -(2n)^2+(2n+1)^2 = (2n+1)(n+1)(2n+3)/3 - 2n(n+1)(2n+1)/3 = (n+1)(2n+1)
do đó ta có khi n = 100 thì
1^2-2^2+3^2-4^2.....+99^2-100^2+101^2 = (100+1)*(2*100+1)=201*101