When proving something by induction, we have to establish a base case: we must prove that our assumption
[tex]P(n):= 10+24+38+.......+(14n-4)=7n^2+3n[/tex]
is true for [tex]n=1[/tex]. We have
[tex](14\cdot 1-4) = 7\cdot 1^2+3\cdot 1 \iff 14-4 =7+3 \iff 10=10[/tex]
With the base case covered, we assume [tex]P(n)[/tex] and prove that if [tex]P(n)[/tex] holds, then [tex]P(n+1)[/tex] follows. We have
[tex]P(n+1):= 10+24+38+.......+(14n-4)+(14(n+1)-4)=7(n+1)^2+3(n+1)[/tex]
Rewrite this expression as
[tex]10+24+38+.......+(14n-4) + 14n+10 = 7n^2+14n+7+3n+3[/tex]
Rearrange the terms as follows:
[tex]10+24+38+.......+(14n-4) + 14n+10 = 7n^2+3n+14n+10[/tex]
We already know (because we are assuming [tex]P(n)[/tex]) that
[tex]10+24+38+.......+(14n-4)=7n^2+3n[/tex]
And when writing [tex]P(n+1)[/tex] we wrote this equation, adding [tex]14n+10[/tex] to both sides.
Recapping, we have assumed that [tex]P(n)[/tex], i.e. we have assumed that
[tex]10+24+38+.......+(14n-4)=7n^2+3n[/tex]
Then we showed that [tex]P(n+1)[/tex] can be written as
[tex]\big(10+24+38+.......+(14n-4)\big)+14n+10=\big(7n^2+3n\big)+14n+10[/tex]
And so, if [tex]P(n)[/tex] is true, [tex]P(n+1)[/tex] must be true as well. This concludes the proof.
