Question

# Solve for every value of n, ${1^2} + {2^2} + {3^2} + ............ + {n^2} = \dfrac{1}{6}\left( n \right)\left( {n + 1} \right)\left( {2n + 1} \right)$, n is a natural number

Hint-Use mathematical induction. It is a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number.

We prove the above relation by principle of mathematical induction. This technique involves two steps to prove a statement.
Step 1-It proves that a statement is true for the initial value.
Step 2-It proves that if the statement is true for the nth number then it is also true for (n+1)th number.
Let $p\left( n \right) = {1^2} + {2^2} + {3^2} + ............ + {n^2} = \dfrac{1}{6}\left( n \right)\left( {n + 1} \right)\left( {2n + 1} \right)$
For n=1
$LHS = {1^2} = 1$
$RHS = \dfrac{{\left( 1 \right)\left( {1 + 1} \right)\left( {2 \times 1 + 1} \right)}}{6} \\ \Rightarrow RHS = \dfrac{{1 \times 2 \times 3}}{6} = 1 \\$
$LHS = RHS$
P(n) is true for n=1
So, It proves that a statement is true for the initial value.
Now, we Assume that P(k) is true
${1^2} + {2^2} + {3^2} + ............ + {\left( {k - 1} \right)^2} + {k^2} = \dfrac{{\left( k \right)\left( {k + 1} \right)\left( {2k + 1} \right)}}{6}$
Now, we have to prove that P(k+1) is true
${1^2} + {2^2} + {3^2} + ............ + {k^2} + {\left( {k + 1} \right)^2} = \dfrac{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {2k + 3} \right)}}{6} \\ LHS = {1^2} + {2^2} + {3^2} + ............ + {k^2} + {\left( {k + 1} \right)^2} \\ \Rightarrow LHS = \dfrac{{\left( k \right)\left( {k + 1} \right)\left( {2k + 1} \right)}}{6} + {\left( {k + 1} \right)^2} \\ \Rightarrow LHS = \dfrac{{\left( k \right)\left( {k + 1} \right)\left( {2k + 1} \right) + 6{{\left( {k + 1} \right)}^2}}}{6} \\$
Take common (k+1)
$\Rightarrow LHS = \dfrac{{\left( {k + 1} \right)\left( {k\left( {2k + 1} \right) + 6\left( {k + 1} \right)} \right)}}{6} \\ \Rightarrow LHS = \dfrac{{\left( {k + 1} \right)\left( {2{k^2} + 7k + 6} \right)}}{6} \\$
Now factories
$\Rightarrow LHS = \dfrac{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {2k + 3} \right)}}{6} \\ LHS = RHS \\$
It is proved that P(k+1) is true. So, we can say P(k) is also true.
So, Hence proved ${1^2} + {2^2} + {3^2} + ............ + {n^2} = \dfrac{1}{6}\left( n \right)\left( {n + 1} \right)\left( {2n + 1} \right)$ for every value of n, n is a natural number.

Note-Whenever we face such types of problems we use some important points. First check the statement for initial value (n=1) .If it is proved then we assume the statement is true for (n=k) .So, statement is also true for (n=k+1).