Question

Tìm n tự nhiên, sao cho

A) 2+4+6+...+2n=210

B)1+3+5+...+2n-1=225

 

Answer 1

\(2+4+6+....+2n=210\)

\(\Rightarrow2\left(1+2+3+.....+n\right)=210\)

\(\Rightarrow1+2+3+...+n=210:2\)

\(\Rightarrow1+2+3+...+n=105\)

\(=\frac{n\left(n+1\right)}{2}=105\)

\(=n\left(n+1\right)=210\)

Vì \(n\left(n+1\right)\) là hai số tự nhiên liên tiếp mà \(210=14.15\)

nên \(n=14\)

1+3+5+...+2n-1=225

\(=\frac{\left(2n-1+1\right)n}{2}=225\)

\(\Rightarrow\frac{2nn}{2}=225\)

\(\frac{2n^2}{2}=225\)

\(=n^2=225\)

Ta có : \(n^2=225=3^2.5^2=15^2\)

\(\Rightarrow n=15\)

Answer 2

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

225 = 1 +3 + 5 +...+ (2n + 1)
= (n + 1)(2n + 1 + 1)/2
= (n + 1)^2
<=> n + 1 = 15
<=> n = 14