document.write( "Question 935544: use mathematical induction to prove the summation of 
\n" ); document.write( "i=1
\n" ); document.write( "4i-1=n(2n+1)
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Algebra.Com's Answer #568805 by KMST(5328)\"\" \"About 
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\n" ); document.write( "\"sum%28%284i-1%29%2C+i=1%2C+i=n+%29=n%282n%2B1%29\"
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\n" ); document.write( "For \"n=1\" , what we need to prove is
\n" ); document.write( "\"sum%28%284i-1%29%2C+i=1%2C+i=1+%29=1%282%2A1%2B1%29\"
\n" ); document.write( "\"sum%28%284i-1%29%2C+i=1%2C+i=1+%29=4%2A1-1=4-1=3\" is the \"sum\" of just the first term, \"3\" .
\n" ); document.write( "\"1%282%2A1%2B1%29=1%282%2B1%29=1%2A3=3\" has the same value.
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\n" ); document.write( "After that we have to prove that
\n" ); document.write( "if \"sum%28%284i-1%29%2C+i=1%2C+i=n+%29=n%282n%2B1%29\" is true for \"n=k\" ,
\n" ); document.write( "it must be true for \"n=k%2B1\" .
\n" ); document.write( "For \"n=k\" ,
\n" ); document.write( "we say it is true that \"sum%28%284i-1%29%2C+i=1%2C+i=n+%29=n%282n%2B1%29\" , so
\n" ); document.write( "\"sum%28%284i-1%29%2C+i=1%2C+i=k+%29=k%282k%2B1%29\" .
\n" ); document.write( "Then, we must prove that it is true for \"n=k%2B1\" , meaning that
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