
If the middle terms in the expansion of \[{({x^2} + \dfrac{1}{x})^{2n}}\] is \[184756{x^{10}}\], then what is the value of \[n\]?
A. \[10\]
B. \[8\]
C. \[5\]
D. \[6\]
Answer
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Hint:To solve this question expand the given expression. And then find the middle term of the given expression and then equate the power with power and the coefficient to the coefficient of the given value in the question. After solving further, find the value of \[n\]. The value expression that is to expand is \[{({x^2} + \dfrac{1}{x})^{2n}}\] this is in the form of \[{(a + b)^n}\].
Complete step-by-step solution:
Given,
An expression to expand \[{({x^2} + \dfrac{1}{x})^{2n}}\]
And the middle term of the expression is \[184756{x^{10}}\].
To find,
The value of \[n\]
Formula used:
Expansion of \[{(a + b)^n}\]
\[{(a + b)^n} = {n_{{C_0}}}{a^n}{b^0} + {n_{{C_1}}}{a^{n - 1}}{b^1} + {n_{{C_2}}}{a^{n - 2}}{b^2} + ................... + {n_{{C_{n - 1}}}}{a^1}{b^{n - 1}} + {n_{{C_n}}}{a^0}{b^n}\]
Here,
In the given question, \[a\] is \[{x^2}\] and \[b\] is \[\dfrac{1}{x}\]
We have to find the middle term in the expansion.
Total power to the expression is \[2n\].
So, the total number of terms is one greater than power. That is \[2n + 1\].
Middle term of the expression in the \[{n^{th}}\] term.
\[{n^{th}}\] term in the expression is
\[{n^{th}}\,term = 2{n_{{C_n}}}{a^{n + 1}}{b^{n - 1}}\]
On putting the value of \[a\] and \[b\] the expression look like
\[{n^{th}}\,term = 2{n_{{C_n}}}{({x^2})^n}(\dfrac{1}{x})n\]
Onn further solving
\[{n^{th}}\,term = 2{n_{{C_n}}}({x^{2n - n}})\]
Now solving the power
\[{n^{th}}\,term = 2{n_{{C_n}}}({x^n})\] …………………………………(i)
This is also middle term
Now from the question middle term is
\[middle\,term = 184756{x^{10}}\] …………………………..……(ii)
On equating the values of middle term from equation 1 and 2
\[2{n_{{C_n}}}({x^n}) = 184756{x^{10}}\]
On comparing the power of x both side we get
\[{x^n} = {x^{10}}\]
From here,
\[n = 10\]
Final answer:
From her value of n satisfying the condition of middle term \[184756{x^{10}}\] is
\[ \Rightarrow n = 10\]
Note: To solve these types of questions you must know the expansion of different terms and after expanding the expression compare the given term of the question with the respective term of the expression. And compare both the terms to get the value of any variable.
Complete step-by-step solution:
Given,
An expression to expand \[{({x^2} + \dfrac{1}{x})^{2n}}\]
And the middle term of the expression is \[184756{x^{10}}\].
To find,
The value of \[n\]
Formula used:
Expansion of \[{(a + b)^n}\]
\[{(a + b)^n} = {n_{{C_0}}}{a^n}{b^0} + {n_{{C_1}}}{a^{n - 1}}{b^1} + {n_{{C_2}}}{a^{n - 2}}{b^2} + ................... + {n_{{C_{n - 1}}}}{a^1}{b^{n - 1}} + {n_{{C_n}}}{a^0}{b^n}\]
Here,
In the given question, \[a\] is \[{x^2}\] and \[b\] is \[\dfrac{1}{x}\]
We have to find the middle term in the expansion.
Total power to the expression is \[2n\].
So, the total number of terms is one greater than power. That is \[2n + 1\].
Middle term of the expression in the \[{n^{th}}\] term.
\[{n^{th}}\] term in the expression is
\[{n^{th}}\,term = 2{n_{{C_n}}}{a^{n + 1}}{b^{n - 1}}\]
On putting the value of \[a\] and \[b\] the expression look like
\[{n^{th}}\,term = 2{n_{{C_n}}}{({x^2})^n}(\dfrac{1}{x})n\]
Onn further solving
\[{n^{th}}\,term = 2{n_{{C_n}}}({x^{2n - n}})\]
Now solving the power
\[{n^{th}}\,term = 2{n_{{C_n}}}({x^n})\] …………………………………(i)
This is also middle term
Now from the question middle term is
\[middle\,term = 184756{x^{10}}\] …………………………..……(ii)
On equating the values of middle term from equation 1 and 2
\[2{n_{{C_n}}}({x^n}) = 184756{x^{10}}\]
On comparing the power of x both side we get
\[{x^n} = {x^{10}}\]
From here,
\[n = 10\]
Final answer:
From her value of n satisfying the condition of middle term \[184756{x^{10}}\] is
\[ \Rightarrow n = 10\]
Note: To solve these types of questions you must know the expansion of different terms and after expanding the expression compare the given term of the question with the respective term of the expression. And compare both the terms to get the value of any variable.
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