Tìm n thuộc N* biết 2+4+6+...............+2.n=210
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: 2+4+6+8+…+2.n=210
=>2.1+2.2+2.3+2.4+…+2.n=210
=>2.(1+2+3+4+…+n)=210
=>1+2+3+4+…+n=210:2
=>n.(n+1):2=210:2
=>n.(n+1)=210=14.15
=>n.(n+1)=14.(14+1)
=>n=14
![](https://rs.olm.vn/images/avt/0.png?1311)
2+4+6+8+...+2x=210
=>2.1+2.2+2.3+2.4+...+2.x=210
=>2.(1+2+3+4+...+x)=210
=>1+2+3+4+...+x=210:2
=>1+2+3+4+...+x=105
=>x=14
2+4+6+8+...+2x=210
=>2.1+2.2+2.3+2.4+...+2.x=210
=>2.(1+2+3+4+...+x)=210
=>1+2+3+4+...+x=210:2
=>1+2+3+4+...+x=105
=>x=14
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 2+4+...+2n=210
=> 2(1+2+...+n)=210
=> \(\frac{2n\left(n+1\right)}{2}=210\)
=> n(n+1) = 210
Mà 14.15 = 210
=> n=14
b, 1+3+....+2n=225
=> \(\frac{\left[\left(2n+1\right)-1\right].n}{2}=225\)
=> \(\frac{2n.n}{2}=225\)
=> n2 = 225
=> \(n=\pm15\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
2 + 4 + 6 + 8 + ... + 2n = 210
=> 2 . (1 + 2 + 3 + 4 + ... + n) = 210
=> 1 + 2 + 3 + 4 + ... + n = 210 : 2
=> 1 + 2 + 3 + 4 + ... + n = 105
=> n . (n + 1) : 2 = 105
=> n . (n + 1) = 105 . 2
=> n . (n + 1) = 210
Vì 14 . 15 = 210 => n = 14
![](https://rs.olm.vn/images/avt/0.png?1311)
2+4+6+8+...+2n = 210
2.(1+2+3+4+...+n) = 210
1+2+3+4+...+n = 210:2
1+2+3+4+...+n = 105
=> n.(n+1):2 = 105
n.(n+1) = 105.2
n.(n+1) = 210
Vì 14.15 = 210 => n = 14
2+4+6+...+2n = 210
((2n-2):2+1). (2n+2):2=210
n.(n+1)= 210
mà 210 = 15.14
=> n+1=15=> n=14
n+1=14=> n= 13(L)
vậy n= 14
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 2 + 4 + 6 + ... + 2n = 210
1.2 + 2.2 + 2.3 + ... + 2n = 210
2.(1+2+3+...+n) = 210
1 + 2 + 3 + n = 105
n(n+1):2 = 105
n(n+1) = 210 = 14.15
=> n = 144
b) 1 + 3 + 5 + ... + ( 2n - 1 ) = 225
( 2n-1 + 1).n/2 = 225
n2 = 225
=> n = 15
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 2 + 4 + 6 + ... + 2n = 210
1.2 + 2.2 + 2.3 + ... + 2n = 210
2.(1+2+3+...+n) = 210
1 + 2 + 3 + ... + n = 105
\(\frac{n\left(n+1\right)}{2}\)= 105
n(n+1) = 210
n(n+1) = 14.15
=> n = 14
b) 1+3+5+...+(2n-1)=225
\(\frac{\left(2n-1+1\right).n}{2}\) =225
\(\frac{2n.n}{2}\) =225
\(\frac{2.n^2}{2}\) =225
\(n^2\) =225
Ta có: \(n^2\) =225 = \(3^2\).\(5^2\)= \(\left(15\right)^2\)
=> n = 15
Mình biết đáp số là 14 nhưng ko biết lời giải các bạn giúp mình nhé
đk: n > 0
210 = 2 + 4 + 6 + ...+ 2n
= n(2 + 2n)/2
= n(1 + n)
= n^2 + n
<=> n^2 + n - 210 = 0
=> n = -15 (loại); n = 14