chung to 2^2004+1 la so chinh phuong
a, so co tong cac chu so la 2004 khong la so chinh phuong
b, so co tong cac chu so la 2006 khong la so chinh phuong
c, n = 20044+20043+20042+23 khong 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
chung minh rang neu a,b la tong cua2 so chinh phuong thi ab cung la tong cua 2 so chinh phuong
S=5^1+5^2+5^3+...+5^2004 cm S la so chinh phuong
cho A=2*3*5*7*9*......*Pn la tich cua aso nguyen to dau tien. Chung to rang 3 so A-1;A;A+1 KHONG CO SO NAO LA SO CHINH PHUONG
cho A =3+3^2+.........+3^2004
chung minh A chia het cho 130
A co phai la so chinh phuong hay khong
lam cu the nhe
cho 1 so chinh phuong co 4 chu so. Biet rang chu so tan cung cua no la so nguyen to, tong cac chu so cua no cung la so chinh phuong va can bac 2 cua no co tong cac chu so cung la so chinh phuong
cho p la tich cua 2016 so nguyen to dau tien
chung minh rang p-1 va p+1 khong la so chinh phuong