Question 1115661
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To prove by mathematical induction:
(1) show the statement is true for some beginning value (usually 1, but not always); and
(2) show that, assuming it is true for some integer n, it follows that it is true for n+1 also<br>
The statement is true for n=1: 5 = (1^2)(2(1)+3) = 1*5 = 5<br>
We need to show that if<br>
5+23+53+...+(6n^2-1) = n^2(2n+3)<br>
is true, then it follows algebraically that<br>
5+23+53+...+(6n^2-1)+(6(n+1)^2-1) = (n+1)^2(2(n+1)+3)<br>
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.<br>
The proof is complete if the two expressions are the same.<br>
On the left...
{{{n^2(2n+3)+(6(n+1)^2-1)}}} =
{{{2n^3+3n^2+6(n^2+2n+1)-1}}} =
{{{2n^3+3n^2+6n^2+12n+6-1}}} =
{{{2n^3+9n^2+12n+5}}}<br>
On the right...
{{{(n+1)^2(2(n+1)+3)}}} =
{{{(n^2+2n+1)(2n+5)}}} =
{{{2n^3+5n^2+4n^2+10n+2n+5}}} =
{{{2n^3+9n^2+12n+5}}}<br>
The proof by mathematical induction is complete.