2021-M2-02

Let $P(n)$ be the given statement.

For $n=1$,

$\begin{array}{cl}& \text{LS}\\ =& {\displaystyle \sum _{k=1}^1(3k^5+k^3)}\\ =& 3(1{)}^5+(1{)}^3\\ =& 4\end{array}$

$\begin{array}{cl}& \text{RS}\\ =& {\displaystyle \frac{1^3(1+1{)}^3}{2}}\\ =& 4\end{array}$

$\therefore \text{LS}=\text{RS}$.

$\therefore P(1)$ is true.

Assume that $P(m)$ is true for some positive integers $m$.

i.e. ${\displaystyle \sum _{k=1}^m(3k^5+k^3)={\displaystyle \frac{m^3(m+1{)}^3}{2}}}$.

For $n=m+1$,

$\begin{array}{cl}& \text{LS}\\ =& {\displaystyle \sum _{k=1}^{m+1}(3k^5+k^3)}\\ =& {\displaystyle \sum _{k=1}^m(3k^5+k^3)+3(m+1{)}^5+(m+1{)}^3}\\ =& {\displaystyle \frac{m^3(m+1{)}^3}{2}}+3(m+1{)}^5+(m+1{)}^3\\ =& {\displaystyle \frac{(m+1{)}^3}{2}}[m^3+6(m+1{)}^2+2]\\ =& {\displaystyle \frac{(m+1{)}^3}{2}}(m^3+6m^2+12m+6+2)\\ =& {\displaystyle \frac{(m+1{)}^3}{2}}(m^3+6m^2+12m+8)\\ =& {\displaystyle \frac{(m+1{)}^3(m+2{)}^3}{2}}\\ =& \text{RS}\end{array}$

$\therefore P(m+1)$ is also true.

Therefore by mathematical induction, $P(n)$ is true for all positive integers $n$.