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# Prove by mathematical induction that ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$.  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.

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.

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