Hi Carol.
Good work so far.
If S(k) > 10k+7, as you've assumed, then
2 S(k) > 2(10k + 7)
but since
2(10k + 7) = 10k + 10 + 7 + 10k  3 = 10(k+1) + 7 + 10k  3
then
2 S(k) > 10(k+1) + 7 + 10k  3
and since k ≥ 10, then 10k  3 > 0, so
10(k+1) + 7 + 10k  3 > 10(k+1) + 7
So if a > b > c then a > c:
2 S(k) > 10(k+1) + 7
and since S(k+1) = 2^(k+1) = 2 2^{k} = 2 S(k), we have proven:
S(k+1) > 10(k+1) + 7.
Stephen La Rocque.>
