document.write( "Question 1115661: use mathematical induction to prove that the statement is true for all positive integers.
\n" ); document.write( "5+23+53+...6n^2 - 1=n^2 (2n+3)
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Algebra.Com's Answer #734075 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "To prove by mathematical induction:
\n" ); document.write( "(1) show the statement is true for some beginning value (usually 1, but not always); and
\n" ); document.write( "(2) show that, assuming it is true for some integer n, it follows that it is true for n+1 also

\n" ); document.write( "The statement is true for n=1: 5 = (1^2)(2(1)+3) = 1*5 = 5

\n" ); document.write( "We need to show that if

\n" ); document.write( "5+23+53+...+(6n^2-1) = n^2(2n+3)

\n" ); document.write( "is true, then it follows algebraically that

\n" ); document.write( "5+23+53+...+(6n^2-1)+(6(n+1)^2-1) = (n+1)^2(2(n+1)+3)

\n" ); document.write( "On the left, replace the sum up to the (6n^2-1) term with the expression n^2(2n+3) and simplify. And simplify the expression on the right.

\n" ); document.write( "The proof is complete if the two expressions are the same.

\n" ); document.write( "On the left...
\n" ); document.write( "\"n%5E2%282n%2B3%29%2B%286%28n%2B1%29%5E2-1%29\" =
\n" ); document.write( "\"2n%5E3%2B3n%5E2%2B6%28n%5E2%2B2n%2B1%29-1\" =
\n" ); document.write( "\"2n%5E3%2B3n%5E2%2B6n%5E2%2B12n%2B6-1\" =
\n" ); document.write( "\"2n%5E3%2B9n%5E2%2B12n%2B5\"

\n" ); document.write( "On the right...
\n" ); document.write( "\"%28n%2B1%29%5E2%282%28n%2B1%29%2B3%29\" =
\n" ); document.write( "\"%28n%5E2%2B2n%2B1%29%282n%2B5%29\" =
\n" ); document.write( "\"2n%5E3%2B5n%5E2%2B4n%5E2%2B10n%2B2n%2B5\" =
\n" ); document.write( "\"2n%5E3%2B9n%5E2%2B12n%2B5\"

\n" ); document.write( "The proof by mathematical induction is complete.
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