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# If the number of terms in ${{\left( 1+{{x}^{-1}}+{{x}^{-2}} \right)}^{n}}$ is 53, then the largest prime p so that $n!$ is divisible by ${{5}^{p-1}}$ is(a) 3(b) 5(c) 7 (d) 11

Last updated date: 20th Jun 2024
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Hint: We solve this problem by using the standard result that is the number of terms in the expansion of ${{\left( A+\dfrac{B}{x}+\dfrac{C}{{{x}^{2}}} \right)}^{n}}$ is given as $2n+1$ then we find the value of $'n'$
After that we find the power of 5 that includes in $n!$ which means the largest power of 5 such that $n!$ is divisible by ${{5}^{p-1}}$ to get the value of ‘p’.

We are given that the expansion of ${{\left( 1+{{x}^{-1}}+{{x}^{-2}} \right)}^{n}}$ has 53 terms.
We know that the number of terms in the expansion of ${{\left( A+\dfrac{B}{x}+\dfrac{C}{{{x}^{2}}} \right)}^{n}}$ is given as $2n+1$
By using the above result to given condition we get
\begin{align} & \Rightarrow 2n+1=53 \\ & \Rightarrow n=26 \\ \end{align}
Now, let us find the multiples of 5 that are involved in $26!$
We know that
$n!=1\times 2\times 3\times 4..........\times n$
By using the above formula we get
$\Rightarrow 26!=1\times 2\times 3\times .........\times 26$
Now, let us write the 5 multiples separately then we get
$\Rightarrow 26!=\left( 5\times 10\times 15\times 20\times 25 \right)\times k$
Here, we can assume that the remaining product as $'k'$
Now let us write the above equation in the power of 5 then we get
$\Rightarrow 26!={{5}^{6}}\left( p \right)$
Here, we can assume that the remaining product as $'p'$
Now, we can see that the highest power of 5 that divides $26!$ exactly is ${{5}^{6}}$
We are given that $n!$ is divisible by ${{5}^{p-1}}$ where $n=26$
Now, by comparing the given statement with the result we get
\begin{align} & \Rightarrow {{5}^{p-1}}={{5}^{6}} \\ & \Rightarrow p-1=6 \\ & \Rightarrow p=7 \\ \end{align}
Therefore the largest prime number ‘p’ such that $n!$ is divisible by ${{5}^{p-1}}$ is 7

So, the correct answer is “Option c”.

Note: Students may make mistakes in taking the formula of number of terms of expansion.
The number of terms in the expansion of ${{\left( A+\dfrac{B}{x}+\dfrac{C}{{{x}^{2}}} \right)}^{n}}$ is given as $2n+1$
We also have other formula that is
The number of terms in the expansion of ${{\left( A+B+C \right)}^{n}}$ is given as $\left( n+1 \right)\left( n+2 \right)$
Students may get confused between these two formulas and get the wrong answer.