document.write( "Question 1179135: Prove 4•10^(2n)+9•10^(2n-1)+5 is divisible by 99 for n is in N. \n" ); document.write( "
Algebra.Com's Answer #808650 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Use mathematical induction.... \n" ); document.write( "(1) show it is true for n=1; then (2) show that if it is true for n=k then it is also true for n=k+1 \n" ); document.write( "(1) When n=1, the expression is 4(100)+9(10)+5 = 495 = 5(99). \n" ); document.write( "The expression is divisible by 99 for n=1. \n" ); document.write( "(2) Assume the expression is true for n=k and show that it follows that it is true for n=k+1: \n" ); document.write( "Assume 4*10^(2k)+9*10(2k-1)+5 is divisible by 99. Then \n" ); document.write( "4*10^(2(k+1))+9*10^(2(k+1)-1)+5 = \n" ); document.write( "4*10^(2k+2)+9*10^(2k+1)+5 = \n" ); document.write( "100(4*10^(2k)+9*10(2k-1)+5)-495 \n" ); document.write( "In that expression, (4*10^(2k)+9*10(2k-1)+5) is divisible by 99 and 495 is also divisible by 99, so we have shown that if the expression is true for n=k then it follows that it is true for n=k+1. \n" ); document.write( "That completes the proof by mathematical induction. \n" ); document.write( " \n" ); document.write( " |