AH la duong cao cua cac hinh tam giac nao?
Viet ten day tuong ung cua hinh tam giac.
\(P=5+5^2+...+5^{101}+5^{102}\)
\(P=5\left(1+5\right)+...+5^{101}\left(1+5\right)\)
\(P=5\cdot6+...+5^{101}\cdot6\)
\(P=6\cdot\left(5+...+5^{101}\right)⋮6\left(đpcm\right)\)
C/m tương tự khi chứng minh chia hết cho 31 ( nhóm 3 số với nhau )
Ta có : \(P=5+5^2+...+5^{102}\)
\(\Rightarrow P=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{101}+5^{102}\right)\)
\(\Rightarrow P=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{101}\left(1+5\right)\)
\(\Rightarrow P=\left(5+5^3+...+5^{101}\right)\left(1+5\right)=6\left(5+5^3+...+5^{101}\right)\)
Ta thấy \(6\left(5+5^3+...+5^{101}\right)⋮6\Rightarrow P⋮6\RightarrowĐPCM\)
Ta lại có : \(P=5+5^2+...+5^{102}\)
\(\Rightarrow P=\left(5+5^2+5^3\right)+...+\left(5^{100}+5^{101}+5^{102}\right)\)
\(\Rightarrow P=5\left(1+5+5^2\right)+...+5^{100}\left(1+5+5^2\right)\)
\(\Rightarrow P=\left(5+...+5^{100}\right)\left(1+5+5^2\right)=31\left(5+...+5^{100}\right)\)
Ta thấy \(31\left(5+...+5^{100}\right)⋮31\Rightarrow P⋮31\RightarrowĐPCM\)
