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)\"\" \"About 
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\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.

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