# 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

Last updated date: 28th Mar 2023

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Answer

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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).

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).

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