CMR:
A=1+3+5+.............+2n-1 (n ∈ N)
A la so chinh phuong
a) cho A = 1+3+5+7+...+(2n+1) Voi n thuoc N
chung to rang A la so chinh phuong
b)B=2+4+6+8+...+2n voi n thuocN
so B co phai la so chinh phuong ko
\(A=1+3+....+\left(2n+1\right)=\frac{\left(2n+2\right)\left(n+1\right)}{2}=\left(n+1\right)^2\)
A = 1 + 3 + 5 + 7 + ... + 2n + 1
= \(\left[\left(2n+1-1\right):2+1\right].\left(\frac{2n+1+1}{2}\right)\)
= \(\left(n+1\right).\left(n+1\right)\)
= \(\left(n+1\right)^2\)
=> A là số chính phương (đpcm)
b) \(2+4+6+...+2n\)
= \(\left[\left(2n-2\right):2+1\right].\frac{2n+2}{2}\)
= \(n.\left(n+1\right)\)
= \(n^2+n\)
\(\Rightarrow\)B không là số chính phương
a) A có số số hạng là: (2n+1-1) :2 +1 = n+1 (số)
=> \(A=\frac{\left(2n+1+1\right).\left(n+1\right)}{2}\)
\(=\frac{2\left(n+1\right)\left(n+1\right)}{2}\)
\(A=\left(n+1\right)^2\)
\(\Rightarrow A\)là số chính phương
a) cho A = 1 + 3 + 5 + 7 +......+(2n + 1) Voi n thuoc N
chung to rang A la so chinh phuong
b) cho B = 2 +4+6 + 8 + ....+ 2n Voi n thuocN
so B co the la chinh phuong ko
Cmr A=1+3+5+...+n la so chinh phuong?
Cmr A=1+3+5+...+n la so chinh phuong?
\(A=\left(1+n\right)\left[\left(n-1\right):2+1\right]:2=\left(\frac{n+1}{2}\right)^2.\)= số chính phưng (n là số lẻ)
chung to rang so A la so chinh phuong biet rang
A = 1+3+5+7+...+(2n-1) voi n € N*
)Cho A= 111...1555...56 ( n chu so 1 , n-1 chu so 5 ) . Cmr : A la so chinh phuong
1+3+5+.......+(2n-1) , n thoc so tu nhien khac 0 . n co phai la so chinh phuong khong
cmr
neu n la so tu nhien n+1 va 2n+1 la so chinh phuong thi n la boi cua 24
giai co loi giai ai dung minh tick
cho n là số tự nhiên khác 0 và a là ước nguyên dương của 2n^2. cmr n^2 + a la so chinh phuong
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là số chính phương.k2<k2+2k<(k+1)2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">