document.write( "Question 389515: Let a_n = 11...1 with 3^n digits. Prove that a_n is divisible by 3a_(n-1). \n" ); document.write( "
Algebra.Com's Answer #276066 by robertb(5830)\"\" \"About 
You can put this solution on YOUR website!
a_1 = 111 = \"111%2A10%5E0+=+111%2A%28%2810%5E3+-+1%29%2F%2810%5E3+-+1%29%29\"
\n" ); document.write( "a_2 = 111 111 111 =
\n" ); document.write( "a_3 = 111 111 111...111 (27 1's) =
\n" ); document.write( "........
\n" ); document.write( "a_n = by induction.\r
\n" ); document.write( "\n" ); document.write( "Then, substituting 999 for \"10%5E3+-+1\".
\n" ); document.write( "\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Now \"%2810%5E%282%2A3%5E%28n-1%29%29+%2B+10%5E%283%5E%28n-1%29%29+%2B+1%29%2F3\" is definitely a positive integer, because the numerator is a positive integer with \"2%2A3%5E%28n-1%29%2B1\" digits, \"2%2A3%5E%28n-1%29+-+2\" of which are 0's, and 3 are 1's. Hence the sum of the digits of the numerator is 3, which makes the number itself divisible by 3.
\n" ); document.write( "
\n" ); document.write( "
\n" );