a, 100 + 99 + 98+…+ 1+ n = 0
b, n + ( n – 1 )+( n – 2 )+….+( n – 100 ) = 0
a, 100 + 99 + 98+…+ 1+ n = 0 b, n + ( n – 1 )+( n – 2 )+….+( n – 100 ) = 0
By Valentina
By Valentina
a, 100 + 99 + 98+…+ 1+ n = 0
b, n + ( n – 1 )+( n – 2 )+….+( n – 100 ) = 0
Đáp án:
Giải thích các bước giải:
`a,100+99+98+…+1+n=0`
`→100+99+98+…+1=-n`
`→(100+1)[(100-1):1+1]:2=-n`
`→10100:2=-n`
`→5050=-n`
`→n=-5050`
`b,n+(n-1)+(n-2)+….+(n-100)=0`
`→n+n-1+n-2+…+n-100=0`
`→101n-(1+2+…+100)=0`
`→101n-{(100+1)[(100-1):1+1]:2}=0`
`→101n-5050=0`
`→101n=5050`
`→n=50`
Em tham khảo:
$a. 100 + 99 + 98+…+ 1+ n = 0$
⇒$\dfrac{(100+1)(100-1+1)}{2}+n=0$
⇒$5050+n=0$
⇒$n=-5050$
$b. n + ( n – 1 )+( n – 2 )+….+( n – 100 ) = 0$
⇒$101n-(1+2+….+100)=0$
⇒$101n-\dfrac{(100+1)(100-1+1)}{2}=0$
⇒$101n-5050=0$
⇒$101n=5050$
⇒$n=50$
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