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a:
Số số hạng trong dãy M là:
(1002-12):10+1=100(số)
=>Sẽ có 50 cặp (1002;992); (982;972);....;(22;12) có hiệu bằng 10
\(M=1002-992+982-972+...+22-12\)
\(=\left(1002-992\right)+\left(982-972\right)+...+\left(22-12\right)\)
\(=10+10+...+10\)
=10*50=500
b: \(N=\left(202+182+...+42+22\right)-\left(192+172+...+32+12\right)\)
\(=\left(202-192\right)+\left(182-172\right)+...+\left(22-12\right)\)
=10+10+...+10
=10*10=100
\(\dfrac{x-1016}{1001}+\dfrac{x-13}{1002}+\dfrac{x+992}{1003}=\dfrac{x+995}{1004}+\dfrac{x-7}{1005}+1\)
<=>\(\dfrac{x-1016}{1001}-1+\dfrac{x-13}{1002}-2+\dfrac{x+992}{1003}-3=\dfrac{x+995}{1004}-3+\dfrac{x-7}{1005}-2\)
<=>\(\dfrac{x-2017}{1001}+\dfrac{x-2017}{1002}+\dfrac{x-2017}{1003}=\dfrac{x-2017}{1004}+\dfrac{x-2017}{1005}\)
<=>\(\left(x-2017\right)\left(\dfrac{1}{1001}+\dfrac{1}{1002}+\dfrac{1}{1003}-\dfrac{1}{1004}-\dfrac{1}{1005}\right)=0\)
vì 1/1001+1/1002+1/1003-1/1004-1/1005 khác 0 nên x-2017=0<=>x=2017
vậy..........
Each term of S is n!(n2 + n + 1) = n![n(n + 1) + 1] = n(n + 1)n! + n!
By definition, n(n + 1)n! + n! = n! + n(n + 1)!
Therefore, S can be simplified as
1! + 1.2! + 2! + 2.3! + ... + 100! + 100.101!
So \(\dfrac{S+1}{101!}=\dfrac{1+1!+1\cdot2!+2!+2\cdot3!+...+100!+100\cdot101!}{101!}\)
\(=\dfrac{2!+1\cdot2!+2!+2\cdot3!+3!+...+100!+100\cdot101!}{101!}\)
\(=\dfrac{3!+2\cdot3!+3!+...+100!+100\cdot101!}{101!}\)
\(=\dfrac{4!+3\cdot4!+4!+...+100!+100\cdot101!}{101!}\)
\(=...\)
\(=\dfrac{100!+99\cdot100!+100!+100\cdot101!}{101!}\)
\(=\dfrac{101!+100\cdot101!}{101!}\)
\(=1+100=101\)
Hence, \(\dfrac{S+1}{101!}=101\)
Ta có
2+4-6-8+10+12-14-16+18+20-22-24+...-2008
=(2+4-6-8)+(10+12-14-16)+(18+20-22-24)+...+(2002+2004-2006-2008)
=(-8)+(-8)+(-8)+...+(-8) (251 số hạng (-8) )
=(-8).251
=-2008