Answer

Verified

484.8k+ views

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).

Recently Updated Pages

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Which one of the following places is not covered by class 10 social science CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

What percentage of the solar systems mass is found class 8 physics CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

How do you graph the function fx 4x class 9 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference Between Plant Cell and Animal Cell

Why is there a time difference of about 5 hours between class 10 social science CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE