Question

Prove that ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}>\dfrac{{{n}^{3}}}{3}$ for all $n\in N$ using principle of mathematical induction.

Answer 1

Hint: In this problem we need to check whether the given condition is correct or wrong by using the principle of mathematical induction. In the principle of mathematical induction, we will follow three steps. In the first step we will assume the given statement as $p\left( n \right)$ and checks whether the statement $p\left( n \right)$ is true or not for $n=1$ by substituting the value $n=1$ in the given statement. In the second step we will assume that the given statement is true for all $n=k$ and write the given statement in terms of $k$ by substituting $n=k$ in $p\left( n \right)$ and this will be considered as our equation one. In the third step we will add ${{\left( k+1 \right)}^{th}}$ term to the both sides of the above equation and simplify the equation. Now we will compare the simplified equation with the given statement in order to check whether the given statement is true for $n=k+1$. In all the three steps if we get the statement as true in all the three steps, then the given equation is true for all $n\in N$.

Complete step by step solution:
Given statement is ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}>\dfrac{{{n}^{3}}}{3}$.
Let us assume the above statement as $p\left( n \right)={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}>\dfrac{{{n}^{3}}}{3}$.
Substituting $n=1$ in the above equation and calculating the value of $p\left( 1 \right)$, then we will get
$\begin{align}
  & p\left( 1 \right)={{1}^{2}}>\dfrac{{{1}^{3}}}{3} \\
 & \Rightarrow p\left( 1 \right)=1>\dfrac{1}{3}\text{ }\left( \text{true} \right) \\
\end{align}$
We have got $p\left( 1 \right)$ as true that means the given statement $p\left( n \right)$ is true for $n=1$.
Let us assume that the given statement $p\left( n \right)$ is true for all $n=k$.
Substituting $n=k$ in the given statement $p\left( n \right)$, then we will have
$p\left( k \right)={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{k}^{2}}>\dfrac{{{k}^{3}}}{3}....\left( \text{i} \right)$

By observing the given statement, we can write the ${{\left( k+1 \right)}^{th}}$ term as ${{\left( k+1 \right)}^{2}}$. Adding this term to both sides of the above equation, then we will get
$p\left( k \right)={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{k}^{2}}+{{\left( k+1 \right)}^{2}}>\dfrac{{{k}^{3}}}{3}+{{\left( k+1 \right)}^{2}}$
Simplifying the right-hand side of the above equation by using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and some basic mathematical operations, then we will have
$\begin{align}
  & p\left( k \right)={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{k}^{2}}+{{\left( k+1 \right)}^{2}}>\dfrac{{{k}^{3}}}{3}+{{k}^{2}}+2k+1 \\
 & \Rightarrow p\left( k \right)={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{k}^{2}}+{{\left( k+1 \right)}^{2}}>\dfrac{1}{3}\left( {{k}^{3}}+3{{k}^{2}}+6k+1 \right) \\
 & \Rightarrow p\left( k \right)={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{k}^{2}}+{{\left( k+1 \right)}^{2}}>\dfrac{1}{3}\left( {{k}^{3}}+3\left( 1 \right){{\left( k \right)}^{2}}+3\left( k \right){{\left( 1 \right)}^{2}}+{{\left( 1 \right)}^{3}}+3k+2 \right) \\
\end{align}$
Using the formula ${{\left( a+b \right)}^{3}}={{a}^{3}}+3a{{b}^{2}}+3{{a}^{2}}b+{{b}^{3}}$ in the above equation, then we will get
$p\left( k \right)={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{k}^{2}}+{{\left( k+1 \right)}^{2}}>\dfrac{{{\left( k+1 \right)}^{3}}}{3}+\dfrac{3k+2}{3}$
Neglecting the term $\dfrac{3k+2}{3}$ in the above equation, then we will have
$p\left( k \right)={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{k}^{2}}+{{\left( k+1 \right)}^{2}}>\dfrac{{{\left( k+1 \right)}^{3}}}{3}$
From the above equation we can say that the given equation is also true for $n=k+1$

Hence the given equation ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}>\dfrac{{{n}^{3}}}{3}$ is proved by using the principle of mathematical induction.

Note: For this problem we can also use the formula or value ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$. Substitute this value in the given equation and follow the same procedure as we have discussed earlier for the principle of mathematical induction.