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![](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
![](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)
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)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](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
![](https://rs.olm.vn/images/avt/0.png?1311)
Đ/S: a, : 14
b.: 15
NÂNG CAO 1 SỐ CHUYÊN ĐỀ LỚP 6
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