Assume its true for n = k
then 8 + 16 + 24+... + 8k = 4k(k + 1)
sum for k+1 terms = 4k(k + 1) + 8(k + 1)
= 4(k + 1)( (k + 2)
= 4(k + 1)( k + 1 + 1)
Now this is the same as the formula for k terms with the k being replaced by k + 1 . So if the formula is correct for k terms then its must also be correct for k+1 terms
Also its true for k = 1 because 8 = 4(1)(1 + 1) so its true for k = 2,3 4 and so
on. - for all values of k ( n).
This compleres the proof.
