a 1 = 1 a n + 1 = a n ∗ 10 n a n = 10 n 2 ( n − 1 ) {\displaystyle {\begin{aligned}&a_{1}=1\\&a_{n+1}=an*10^{n}\\&a_{n}=10^{{\frac {n}{2}}(n-1)}\\\end{aligned}}}
a n = 10 n 2 ( n − 1 ) = 10 1 2 ( 1 − 1 ) = 10 0 = 1 a 1 = 1 1 = 1 √ {\displaystyle {\begin{aligned}&a_{n}=10^{{\frac {n}{2}}(n-1)}=10^{{\frac {1}{2}}(1-1)}=10^{0}=1\\&a_{1}=1\\&1=1\surd \\\end{aligned}}}
a k = 10 k 2 ( k − 1 ) {\displaystyle \ a_{k}=10^{{\frac {k}{2}}(k-1)}}
a k + 1 = 10 k + 1 2 ( k ) {\displaystyle {\begin{aligned}&a_{k+1}=10^{{\frac {k+1}{2}}(k)}\\\end{aligned}}}
a k + 1 = a k ⏟ 10 k 2 ( k − 1 ) ∗ 10 k a k + 1 = 10 k 2 ( k − 1 ) ∗ 10 k a k + 1 = 10 k 2 ( k − 1 ) + k a k + 1 = 10 k ( k − 1 ) + 2 k 2 a k + 1 = 10 k 2 − k + 2 k 2 a k + 1 = 10 k 2 + k 2 a k + 1 = 10 k ( k + 1 ) 2 {\displaystyle {\begin{aligned}&a_{k+1}=\underbrace {a_{k}} _{10^{{\frac {k}{2}}(k-1)}}*10^{k}\\&a_{k+1}=10^{{\frac {k}{2}}(k-1)}*10^{k}\\&a_{k+1}=10^{{\frac {k}{2}}(k-1)+k}\\&a_{k+1}=10^{\frac {k(k-1)+2k}{2}}\\&a_{k+1}=10^{\frac {k^{2}-k+2k}{2}}\\&a_{k+1}=10^{\frac {k^{2}+k}{2}}\\&a_{k+1}=10^{\frac {k(k+1)}{2}}\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.
טבעי
