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Hint: First of all, put n = 1, 2 and 3 in a given sequence to show that the first three terms of the sequence are zero. Then split the given sequence into its factors to show that all other terms are positive, that is for n > 3, the given sequence is positive.

Here, we are given a sequence defined by \[{{a}_{n}}={{n}^{3}}-6{{n}^{2}}+11n-6\].

We have to show that the first three terms of the sequence are zero and all other terms are positive.

Let us consider the sequence given in the question,

\[{{a}_{n}}={{n}^{3}}-6{{n}^{2}}+11n-6.....\left( i \right)\]

Now, to get the first term of this sequence, we will put n = 1 in equation (i), we will get,

\[{{a}_{1}}={{1}^{3}}-6{{\left( 1 \right)}^{2}}+11\left( 1 \right)-6\]

By simplifying the above equation, we get,

\[\Rightarrow {{a}_{1}}=1-6+11-6\]

\[\Rightarrow {{a}_{1}}=12-12\]

Therefore, we get \[{{a}_{1}}=0\].

Hence, we have proved that the first term of this sequence is 0.

Now, to get the second term of this sequence, we will put n = 2 in equation (i), we will get,

\[{{a}_{2}}={{\left( 2 \right)}^{3}}-6{{\left( 2 \right)}^{2}}+11\left( 2 \right)-6\]

By simplifying the above equation, we get,

\[{{a}_{2}}=8-6\left( 4 \right)+22-6\]

\[\Rightarrow {{a}_{2}}=8-24+22-6\]

\[\Rightarrow {{a}_{2}}=30-30\]

Therefore, we get \[{{a}_{2}}=0\].

Hence, we have proved that the second term of this sequence is 0.

Now, to get the third term of this sequence, we will put n = 3 in equation (i), we will get

\[{{a}_{3}}={{\left( 3 \right)}^{3}}-6{{\left( 3 \right)}^{2}}+11\left( 3 \right)-6\]

By simplifying the above equation, we get

\[{{a}_{3}}=27-6\left( 9 \right)+33-6\]

\[\Rightarrow {{a}_{3}}=27-54+33-6\]

\[\Rightarrow {{a}_{3}}=60-60\]

Therefore, we get \[{{a}_{3}}=0\].

Hence, we have proved that the third term of this sequence is 0.

We have found that,

\[\begin{align}

& {{a}_{1}}=0 \\

& {{a}_{2}}=0 \\

& {{a}_{3}}=0 \\

\end{align}\]

Thus, the given sequence \[{{a}_{n}}={{n}^{3}}-6{{n}^{2}}+11n-6\] is equal to zero for n = 1, 2, 3.

Therefore, we can say that (n – 1), (n – 2) and (n – 3) are factors of the given sequence.

As the given sequence is a cubic equation or 3 degree equation in ‘n’. Therefore, it will have 3 factors. We have already found these factors as ( n – 1), (n – 2) and (n – 3).

Therefore, we can write the given sequence in the terms of its factors as,

\[{{a}_{n}}={{n}^{3}}-6{{n}^{2}}+11n-6=\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)\]

Now, for any value of n greater than 3, we get,

\[\begin{align}

& \left( n-1 \right)>0 \\

& \left( n-2 \right)>0 \\

& \left( n-3 \right)>0 \\

\end{align}\]

For instance, take n = 5, then we get,

As, (n – 1), (n – 2), (n – 3) is greater than 0 for n > 3. Therefore, we get \[{{a}_{n}}=\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)>0\] for n > 3.

For instance, if we take n = 5, we get,

\[{{a}_{n}}=\left( 5-1 \right)\left( 5-2 \right)\left( 5-3 \right)\]

\[{{a}_{n}}=4\times 3\times 2\]

\[{{a}_{n}}=24\] which is greater than 0.

We know that, to get other terms of sequence apart from first, second and third, we have to put n > 3 like n = 4, 5, 6….. in the given sequence.

As we have proved that \[{{a}_{n}}=0\] for n =1, 2, 3 and \[{{a}_{n}}>0\] for n > 3 that is n = 4, 5, 6….so on. Therefore, we can say that the first three terms of the sequence are zero and all other terms are positive.

Note: Here students must note that if we have a cubic equation \[a{{x}^{3}}+b{{x}^{2}}+cx+d\] and its zeroes are \[\alpha ,\beta \] and \[\gamma \], that is, the equation becomes zero at \[x=\alpha ,\text{ }x=\beta \] and \[x=\gamma \]. Then we can write the equation as \[a{{x}^{3}}+b{{x}^{2}}+cx+d=a\left( x-\alpha \right)\left( x-\beta \right)\left( x-\gamma \right)=0\]. In the above question, we had a coefficient of \[{{x}^{3}}={{n}^{3}}\], that is a = 1.

Here, we are given a sequence defined by \[{{a}_{n}}={{n}^{3}}-6{{n}^{2}}+11n-6\].

We have to show that the first three terms of the sequence are zero and all other terms are positive.

Let us consider the sequence given in the question,

\[{{a}_{n}}={{n}^{3}}-6{{n}^{2}}+11n-6.....\left( i \right)\]

Now, to get the first term of this sequence, we will put n = 1 in equation (i), we will get,

\[{{a}_{1}}={{1}^{3}}-6{{\left( 1 \right)}^{2}}+11\left( 1 \right)-6\]

By simplifying the above equation, we get,

\[\Rightarrow {{a}_{1}}=1-6+11-6\]

\[\Rightarrow {{a}_{1}}=12-12\]

Therefore, we get \[{{a}_{1}}=0\].

Hence, we have proved that the first term of this sequence is 0.

Now, to get the second term of this sequence, we will put n = 2 in equation (i), we will get,

\[{{a}_{2}}={{\left( 2 \right)}^{3}}-6{{\left( 2 \right)}^{2}}+11\left( 2 \right)-6\]

By simplifying the above equation, we get,

\[{{a}_{2}}=8-6\left( 4 \right)+22-6\]

\[\Rightarrow {{a}_{2}}=8-24+22-6\]

\[\Rightarrow {{a}_{2}}=30-30\]

Therefore, we get \[{{a}_{2}}=0\].

Hence, we have proved that the second term of this sequence is 0.

Now, to get the third term of this sequence, we will put n = 3 in equation (i), we will get

\[{{a}_{3}}={{\left( 3 \right)}^{3}}-6{{\left( 3 \right)}^{2}}+11\left( 3 \right)-6\]

By simplifying the above equation, we get

\[{{a}_{3}}=27-6\left( 9 \right)+33-6\]

\[\Rightarrow {{a}_{3}}=27-54+33-6\]

\[\Rightarrow {{a}_{3}}=60-60\]

Therefore, we get \[{{a}_{3}}=0\].

Hence, we have proved that the third term of this sequence is 0.

We have found that,

\[\begin{align}

& {{a}_{1}}=0 \\

& {{a}_{2}}=0 \\

& {{a}_{3}}=0 \\

\end{align}\]

Thus, the given sequence \[{{a}_{n}}={{n}^{3}}-6{{n}^{2}}+11n-6\] is equal to zero for n = 1, 2, 3.

Therefore, we can say that (n – 1), (n – 2) and (n – 3) are factors of the given sequence.

As the given sequence is a cubic equation or 3 degree equation in ‘n’. Therefore, it will have 3 factors. We have already found these factors as ( n – 1), (n – 2) and (n – 3).

Therefore, we can write the given sequence in the terms of its factors as,

\[{{a}_{n}}={{n}^{3}}-6{{n}^{2}}+11n-6=\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)\]

Now, for any value of n greater than 3, we get,

\[\begin{align}

& \left( n-1 \right)>0 \\

& \left( n-2 \right)>0 \\

& \left( n-3 \right)>0 \\

\end{align}\]

For instance, take n = 5, then we get,

As, (n – 1), (n – 2), (n – 3) is greater than 0 for n > 3. Therefore, we get \[{{a}_{n}}=\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)>0\] for n > 3.

For instance, if we take n = 5, we get,

\[{{a}_{n}}=\left( 5-1 \right)\left( 5-2 \right)\left( 5-3 \right)\]

\[{{a}_{n}}=4\times 3\times 2\]

\[{{a}_{n}}=24\] which is greater than 0.

We know that, to get other terms of sequence apart from first, second and third, we have to put n > 3 like n = 4, 5, 6….. in the given sequence.

As we have proved that \[{{a}_{n}}=0\] for n =1, 2, 3 and \[{{a}_{n}}>0\] for n > 3 that is n = 4, 5, 6….so on. Therefore, we can say that the first three terms of the sequence are zero and all other terms are positive.

Note: Here students must note that if we have a cubic equation \[a{{x}^{3}}+b{{x}^{2}}+cx+d\] and its zeroes are \[\alpha ,\beta \] and \[\gamma \], that is, the equation becomes zero at \[x=\alpha ,\text{ }x=\beta \] and \[x=\gamma \]. Then we can write the equation as \[a{{x}^{3}}+b{{x}^{2}}+cx+d=a\left( x-\alpha \right)\left( x-\beta \right)\left( x-\gamma \right)=0\]. In the above question, we had a coefficient of \[{{x}^{3}}={{n}^{3}}\], that is a = 1.

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