Q6Discrete Mathematics Structure

Step 1: Base case $n=1$. LHS=$1^3=1$, RHS=$[1(2)/2{]}^2=1$. True.

Step 2: Induction hypothesis. Assume $S(k)$ is true: $1^3+\cdots +k^3=[k(k+1)/2{]}^2$.

Step 3: Prove $S(k+1)$. $1^3+\cdots +k^3+(k+1{)}^3=[k(k+1)/2{]}^2+(k+1{)}^3=[\frac{k(k+1)}{2}{]}^2+(k+1{)}^3$

$=(k+1{)}^2[\frac{k^2}{4}+(k+1)]=(k+1{)}^2[\frac{k^2+4k+4}{4}]=(k+1{)}^2[\frac{(k+2{)}^2}{4}]=[\frac{(k+1)(k+2)}{2}{]}^2$. Proven.