a. Base case : n=1
S(n)= $\frac{1}{2}$$\times$(1)$\times$(1+1)=1.
$\therefore$ Our hypothesis is true for n=1.
Let this hypothesis be true for all integers $\leq$ n. ,i.e S(n)=$\frac{n \times (n+1)}{2}$=1+2+...+n
Then S(n+1)=S(n)+n+1 [From definition]
$\Rightarrow$ S(n)=$\frac{n \times (n+1)}{2}$+$\frac{n \times (n+1)}{2}+(n+1)=(n+1)(\frac{n}{2}+1)=\frac{1}{2} \times (n+1)(n+2)=\frac{1}{2}(n+1)(n+1+1)$=S(n+1).
$\because$ Our theorem is true for n+1, our theorem is true for all integers.
b. Base case : n=1
$\frac{1}{4}$$\times$($n^{4}+2n^3+n^2$)= $\frac{1}{4}$(1+2+1)=1=C(1).
Our hypothesis is true for n=1.
Let this hypothesis be true for all integers $\leq$ n. ,i.e C(n)=$\frac{n^{4}+2n^3+n^2}{4}$=$1^3+2^3+...+n^3$
C(n+1)=C(n)+$(n+1)^3$=$\frac{n^{4}+2n^3+n^2}{4}$+$(n+1)^3$=$\frac{(n+1)^{4}+2(n+1)^3+(n+1)^2}{4}$
$\because$ our Hypothesis is true for n+1 , our theorem is true for all integers $\geq$0.
