
The sum of the third from the beginning and the third from the end of the binomial coefficients in the expression of .${\left( {\sqrt[4]{3} + \sqrt[3]{4}} \right)^n}$ is equal to 9900. The number of rational terms contained in the expansion is
A.8
B.9
C.10
D.11
Answer
513k+ views
Hint: Find the third term, both from beginning and end using the combination. The value of n then use it to find integers and then the rational number through the divisibility condition. A binomial is a polynomial with two terms.
Complete step-by-step answer:
Given that:
In the expansion of \[{(\sqrt[4]{3} + \sqrt[3]{4})^n}\]
We have,
Sum of third from beginning and the third from last = 9900.
Therefore
\[\begin{array}{*{20}{c}}
{^n{C_2}}& + &{^n{C_{n-2}} = {\text{ }}9900} \\
\downarrow &{}& \downarrow \\
{Third{\text{ }}term{\text{ }}from{\text{ }}beginning}&{}&{third{\text{ }}term{\text{ }}from{\text{ }}end}
\end{array}\]
Also, these 2 terms will be equal as we know in combination 3rd from last and 3rd from beginning are equal.
So, \[^n{C_2}{ = ^n}{C_n}_{-{\text{ }}2}\]
\[{ \Rightarrow ^n}{C_2} = {\text{ }}4950\]
\[ \Rightarrow \dfrac{{n(n - 1)}}{2} = 4950\]--(1)
On solving this equation is (1)
\[{n^2}-{\text{ }}100n + {\text{ }}99n-{\text{ }}9900{\text{ }} = {\text{ }}0\]
\[ \Rightarrow n\left( {n-{\text{ }}100} \right){\text{ }} + {\text{ }}99{\text{ }}\left( {n-{\text{ }}100} \right){\text{ }} = {\text{ }}0\]
\[ \Rightarrow (n + {\text{ }}99){\text{ }}\left( {n-{\text{ }}100} \right){\text{ }} = {\text{ }}0\]
∴ n = 100 (n cannot be negative)
Now, we need to find the rational terms.
Therefore,\[\dfrac{r}{2}\] and \[\dfrac{{100 - r}}{3}\] must be integers.
For the second condition,
r = 1, 4, 7, 10, 13, 16,.... 99
r must also be divisible by 4.
Hence, r can take the values:
4, 16, 28, 40, 52, 64, 76, 88, 100 (9 terms).
Therefore, there are 9 terms which are rational.
Hence B option is the correct option.
Note: In this type of questions, we need to know about the calculation of each term and about the binomial expansion. In elementary algebra, the binomial theorem describes the algebraic expansion of powers of a binomial. The Binomial theorem tells us how to expand expressions of the form\[{\left( {a + b} \right)^n}\], for example, \[{\left( {x + y} \right)^7}.\] The larger the power is, the harder it is to expand expressions like this directly. But with the Binomial theorem, the process is relatively fast.
Complete step-by-step answer:
Given that:
In the expansion of \[{(\sqrt[4]{3} + \sqrt[3]{4})^n}\]
We have,
Sum of third from beginning and the third from last = 9900.
Therefore
\[\begin{array}{*{20}{c}}
{^n{C_2}}& + &{^n{C_{n-2}} = {\text{ }}9900} \\
\downarrow &{}& \downarrow \\
{Third{\text{ }}term{\text{ }}from{\text{ }}beginning}&{}&{third{\text{ }}term{\text{ }}from{\text{ }}end}
\end{array}\]
Also, these 2 terms will be equal as we know in combination 3rd from last and 3rd from beginning are equal.
So, \[^n{C_2}{ = ^n}{C_n}_{-{\text{ }}2}\]
\[{ \Rightarrow ^n}{C_2} = {\text{ }}4950\]
\[ \Rightarrow \dfrac{{n(n - 1)}}{2} = 4950\]--(1)
On solving this equation is (1)
\[{n^2}-{\text{ }}100n + {\text{ }}99n-{\text{ }}9900{\text{ }} = {\text{ }}0\]
\[ \Rightarrow n\left( {n-{\text{ }}100} \right){\text{ }} + {\text{ }}99{\text{ }}\left( {n-{\text{ }}100} \right){\text{ }} = {\text{ }}0\]
\[ \Rightarrow (n + {\text{ }}99){\text{ }}\left( {n-{\text{ }}100} \right){\text{ }} = {\text{ }}0\]
∴ n = 100 (n cannot be negative)
Now, we need to find the rational terms.
Therefore,\[\dfrac{r}{2}\] and \[\dfrac{{100 - r}}{3}\] must be integers.
For the second condition,
r = 1, 4, 7, 10, 13, 16,.... 99
r must also be divisible by 4.
Hence, r can take the values:
4, 16, 28, 40, 52, 64, 76, 88, 100 (9 terms).
Therefore, there are 9 terms which are rational.
Hence B option is the correct option.
Note: In this type of questions, we need to know about the calculation of each term and about the binomial expansion. In elementary algebra, the binomial theorem describes the algebraic expansion of powers of a binomial. The Binomial theorem tells us how to expand expressions of the form\[{\left( {a + b} \right)^n}\], for example, \[{\left( {x + y} \right)^7}.\] The larger the power is, the harder it is to expand expressions like this directly. But with the Binomial theorem, the process is relatively fast.
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