Question 935544
You probably meant
{{{sum((4i-1), i=1, i=n )=n(2n+1)}}}
 
For {{{n=1}}} , what we need to prove is
{{{sum((4i-1), i=1, i=1 )=1(2*1+1)}}}
{{{sum((4i-1), i=1, i=1 )=4*1-1=4-1=3}}} is the "sum" of just the first term, {{{3}}} .
{{{1(2*1+1)=1(2+1)=1*3=3}}} has the same value.
 
After that we have to prove that
if {{{sum((4i-1), i=1, i=n )=n(2n+1)}}} is true for {{{n=k}}} ,
it must be true for {{{n=k+1}}} .
For {{{n=k}}} ,
we say it is true that {{{sum((4i-1), i=1, i=n )=n(2n+1)}}} , so
{{{sum((4i-1), i=1, i=k )=k(2k+1)}}} .
Then, we must prove that it is true for {{{n=k+1}}} , meaning that
{{{sum((4i-1), i=1, i=k+1 )=(k+1)(2(k+1)+1)=(k+1)(2k+2+1)=(k+1)(2k+3)}}}
{{{sum((4i-1), i=1, i=k+1 )=sum((4i-1), i=1, i=k )+sum((4i-1), i=k+1, i=k+1 )=k(2k+1)+(4(k+1)-1) =2k^2+k+(4k+4-1)=2k^2+k+(4k+3)=2k^2+5k+3=2k^2+2k+3k+3=2k(k+1)+3(k+1)=(k+1)(2k+3)}}}