1)for n = 24, 25 , 26, 27 (we explicitly show how these can be achieved)
24 = 2 * 5 + 2 * 7
25 = 5 * 5
26 = 1 * 5 + 3 * 7
27 = 4 * 5 + 1 * 7
Let's assume that the statement holds for all k such that n ≥ k ≥ 24.
Then for n + 1 cents postage case, if (n + 1) - 5 = n - 4 ≥ 24 , then by inductive assumption, (n-4) cents can be achieved by using only 5-cents and 7-cents stamps, and hence (n+1) cents can be achieved by using one extra 5-cent stamp as in (n-4) cents case. if (n-4) < 24 then n must be one of those values : 24 , 25, 25 or 27 (since n ≥ 24 ), and it has been justified in basic step. Hence the statement is true for all n ≥ 24.
2) a) c2=c1+22=4
c3=c1+32=9
c4=c2+42=20
c5=c2+52=29
b)"C_1=0<4, C_2=C_1+1=1<16"
if we suppose "C_k<4k^2" for k≥2, then:
For k=2n+1 we have
"C_{k+1}=C_n+4n^2<8n^2=2(k+1)^2<4(k+1)^2;"
For k=2n we have
"C_{k+1}=C_n+(2n+1)^2<4n^2+(2n+1)^2=2k^2+2k+1<4(k+1)^2."
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