Answer
Verified
417.3k+ views
Hint: Using mathematical induction, we show that the equation is true by putting the value of 1 in both L.H.S and R.H.S. Then we need to make the equation true for value. If it is true for this value, then it is true for the next value, which makes it true for all integers greater than that value as well.
Complete step-by-step answer:
Before proceeding with the question, we must know how to use mathematical induction. In mathematical induction, while proving equations, we have to make the equation true by putting the value 1 on both L.H.S and R.H.S. Then we need to make the equation true for a value. Once we prove that it is true for this value, then it will be true for the next value, which will make it true for all integers greater than that value as well.
In this question, we have to prove that ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ where $n$ is the number of terms, using mathematical induction.
So, first, we need to show that it is true for $n=1$.
$\therefore L.H.S=1$
$\Rightarrow R.H.S=\dfrac{1\left( 1+1 \right)\left( 2+1 \right)}{6}$
$\Rightarrow R.H.S=\dfrac{1\times 2\times 3}{6}$
$\therefore R.H.S=1$
$\because L.H.S=R.H.S$
Therefore, the equation is true for $n=1$.
Now, we need to show if it is true for a value, then it is true for the next value, then it is true for all next values.
Let us put $n=n+1$. Therefore, adding ${{\left( n+1 \right)}^{2}}$ on both the sides we get,
$\Rightarrow {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+......+{{n}^{2}}+{{\left( n+1 \right)}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+{{\left( n+1 \right)}^{2}}$
On the R.H.S we get:
$\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+{{\left( n+1 \right)}^{2}}$
$\Rightarrow \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{6{{\left( n+1 \right)}^{2}}}{6}$
$\Rightarrow \dfrac{n\left( n+1 \right)\left( 2n+1 \right)+6{{\left( n+1 \right)}^{2}}}{6}$
Taking $\left( n+1 \right)$ common we get:
$\Rightarrow \dfrac{n+1\left( n\left( 2n+1 \right)+6\left( n+1 \right) \right)}{6}$
$\Rightarrow \dfrac{\left( n+1 \right)\left( n\times 2n+n+6n+6 \right)}{6}$
$\Rightarrow \dfrac{\left( n+1 \right)\left( 2{{n}^{2}}+7n+6 \right)}{6}$
Use middle term splitting to find out the factors of $2{{n}^{2}}+7n+6$, we get,$\Rightarrow \dfrac{\left( n+1 \right)\left( 2{{n}^{2}}+4n+3n+6 \right)}{6}$
$\Rightarrow \dfrac{\left( n+1 \right)\left( 2n\left( n+2 \right)+3\left( n+2 \right) \right)}{6}$
$\Rightarrow \dfrac{\left( n+1 \right)\left( 2n+3 \right)\left( n+2 \right)}{6}$
We can write $2n+3$ as $\left[ 2\left( n+1 \right)+1 \right]$ and $n+2$ as $\left[ \left( n+1 \right)+1 \right]$.
$\Rightarrow \dfrac{\left( n+1 \right)\left[ 2\left( n+1 \right)+1 \right]\left[ \left( n+1 \right)+1 \right]}{6}$
This result is the same if we put $\left( n+1 \right)$in place of $n$. So, it is true for all values for n and values greater than $n$. Hence, proved.
Note: We must be very careful while doing the calculation. For the solution to be easy, try to factorise into factors wherever possible. Try to convert terms in the way we need to prove, for example, we can write $2n+3$ as $\left[ 2\left( n+1 \right)+1 \right]$ and $n+2$ as $\left[ \left( n+1 \right)+1 \right]$. The factors of the quadratic equation can be found using any method, but the middle term split method is the best in this case.
Complete step-by-step answer:
Before proceeding with the question, we must know how to use mathematical induction. In mathematical induction, while proving equations, we have to make the equation true by putting the value 1 on both L.H.S and R.H.S. Then we need to make the equation true for a value. Once we prove that it is true for this value, then it will be true for the next value, which will make it true for all integers greater than that value as well.
In this question, we have to prove that ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ where $n$ is the number of terms, using mathematical induction.
So, first, we need to show that it is true for $n=1$.
$\therefore L.H.S=1$
$\Rightarrow R.H.S=\dfrac{1\left( 1+1 \right)\left( 2+1 \right)}{6}$
$\Rightarrow R.H.S=\dfrac{1\times 2\times 3}{6}$
$\therefore R.H.S=1$
$\because L.H.S=R.H.S$
Therefore, the equation is true for $n=1$.
Now, we need to show if it is true for a value, then it is true for the next value, then it is true for all next values.
Let us put $n=n+1$. Therefore, adding ${{\left( n+1 \right)}^{2}}$ on both the sides we get,
$\Rightarrow {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+......+{{n}^{2}}+{{\left( n+1 \right)}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+{{\left( n+1 \right)}^{2}}$
On the R.H.S we get:
$\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+{{\left( n+1 \right)}^{2}}$
$\Rightarrow \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{6{{\left( n+1 \right)}^{2}}}{6}$
$\Rightarrow \dfrac{n\left( n+1 \right)\left( 2n+1 \right)+6{{\left( n+1 \right)}^{2}}}{6}$
Taking $\left( n+1 \right)$ common we get:
$\Rightarrow \dfrac{n+1\left( n\left( 2n+1 \right)+6\left( n+1 \right) \right)}{6}$
$\Rightarrow \dfrac{\left( n+1 \right)\left( n\times 2n+n+6n+6 \right)}{6}$
$\Rightarrow \dfrac{\left( n+1 \right)\left( 2{{n}^{2}}+7n+6 \right)}{6}$
Use middle term splitting to find out the factors of $2{{n}^{2}}+7n+6$, we get,$\Rightarrow \dfrac{\left( n+1 \right)\left( 2{{n}^{2}}+4n+3n+6 \right)}{6}$
$\Rightarrow \dfrac{\left( n+1 \right)\left( 2n\left( n+2 \right)+3\left( n+2 \right) \right)}{6}$
$\Rightarrow \dfrac{\left( n+1 \right)\left( 2n+3 \right)\left( n+2 \right)}{6}$
We can write $2n+3$ as $\left[ 2\left( n+1 \right)+1 \right]$ and $n+2$ as $\left[ \left( n+1 \right)+1 \right]$.
$\Rightarrow \dfrac{\left( n+1 \right)\left[ 2\left( n+1 \right)+1 \right]\left[ \left( n+1 \right)+1 \right]}{6}$
This result is the same if we put $\left( n+1 \right)$in place of $n$. So, it is true for all values for n and values greater than $n$. Hence, proved.
Note: We must be very careful while doing the calculation. For the solution to be easy, try to factorise into factors wherever possible. Try to convert terms in the way we need to prove, for example, we can write $2n+3$ as $\left[ 2\left( n+1 \right)+1 \right]$ and $n+2$ as $\left[ \left( n+1 \right)+1 \right]$. The factors of the quadratic equation can be found using any method, but the middle term split method is the best in this case.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Two charges are placed at a certain distance apart class 12 physics CBSE
Difference Between Plant Cell and Animal Cell
What organs are located on the left side of your body class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The planet nearest to earth is A Mercury B Venus C class 6 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is BLO What is the full form of BLO class 8 social science CBSE