Question 1137605
Base case:  n=1
LHS: 3
RHS: 1(2*1+1) = 1(3) = 3
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Hypothesis: 3+7+11+...+(4k-1) = k(2k+1)
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Must show the above is true for n=k+1

3+7+11+...+(4k-1)+(4(k+1)-1) =
{{{ green(3+7+11) }}} {{{ green(matrix(1,3,"+","...","+"))}}} {{{ green((4k-1)) }}} + (4(k+1)-1)  

By hypothesis, the green terms are  k(2k+1), so we write
= {{{ k(2k+1) + (4(k+1)-1) }}}
= {{{(2k^2 + k) + (4k+3) }}}
= {{{ 2k^2 + 5k + 3 }}}
= {{{ (k+1)(2k+3) }}}
(n=k+1)
= {{{   n(2n+1) }}}   DONE