document.write( "Question 170325: Use mathematical induction to prove: 4^n+1 + 5^2n-1 is divisible by 21 \n" ); document.write( "
Algebra.Com's Answer #125771 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
Note: I'm assuming that the assumption holds for \"n%3E=1\"\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Step 1) \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Prove for n=1\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " which is divisible by 21\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "-----------------------------------------------\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Step 2) \r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Assume \"4%5E%28k%2B1%29+%2B+5%5E%282k-1%29\" is divisible by 21 (ie assume kth term is divisible by 21)\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "-----------------------------------------------\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Step 3) \r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Prove true for k+1 term\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"4%5E%28k%2B1%29+%2B+5%5E%282k-1%29\" Start with the assumed portion\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"4%5E%28k%2B1%2B1%29+%2B+5%5E%282%28k%2B1%29-1%29\" Plug in k+1 for every k\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"4%5E%28k%2B1%2B1%29+%2B+5%5E%282k%2B2-1%29\" Distribute\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"4%2A4%5E%28k%2B1%29+%2B+5%5E2%2A5%5E%282k-1%29\" Break up the exponent\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"4%2A4%5E%28k%2B1%29+%2B+25%2A5%5E%282k-1%29\" Square 5 to get 25\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"4%2A4%5E%28k%2B1%29+%2B+4%2A5%5E%282k-1%29%2B21%2A5%5E%282k-1%29\" Break up 25 to get 4+21\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"4%284%5E%28k%2B1%29+%2B+5%5E%282k-1%29%29%2B21%2A5%5E%282k-1%29\" Factor out the GCF 4\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Since we're assuming that \"4%5E%28k%2B1%29+%2B+5%5E%282k-1%29\" is divisible by 21, this means that \"4%5E%28k%2B1%29+%2B+5%5E%282k-1%29=21m\" for some integer \"m\"\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"4%2821m%29%2B21%2A5%5E%282k-1%29\" Replace \"4%5E%28k%2B1%29+%2B+5%5E%282k-1%29\" with 21m\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Now let \"n=5%5E%282k-1%29\" (which is an integer)\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"4%2821m%29%2B21n\" Replace \"5%5E%282k-1%29\" with \"n\"\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"21%284m%2Bn%29\" Factor out the GCF 21\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Now let \"j=4m%2Bn\" so the expression becomes\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"21j\"\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Since 21 is a factor of \"21j\", this shows that \"4%5E%28k%2B1%2B1%29+%2B+5%5E%282%28k%2B1%29-1%29\" is divisible by 21. \r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "So this proves that \"4%5E%28n%2B1%29+%2B+5%5E%282n-1%29\" is divisible by 21 for \"n%3E=1\"\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );