
The sum of the coefficient in the expansion of $ {\left( {x + y} \right)^n} $ is 4096. The greatest coefficient in the expansion is-
A.1024
B.924
C.824
D.724
Answer
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Hint: In this question, we need to determine the greatest coefficient present in the expansion of $ {\left( {x + y} \right)^n} $ such that the sum of all the coefficients present in the expansion is 4096. For this, we will use the binomial expansion method.
Complete step-by-step answer:
Binomial expansion theorem is a theorem which specifies the expansion of any power \[{\left( {a + b} \right)^n}\] of a binomial \[\left( {a + b} \right)\] as a sum of products e.g. \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]. The number of in the expansion depends upon the raised positive integral power. If the raised power of expansion is\[n\] then the number of terms of the expansion will be \[n + 1\].
The standard expansion of the Binomial theorem has been given as
$ {\left( {a + b} \right)^n} = {C_0}{a^n} + {C_1}{a^{n - 1}}b + {C_2}{a^{n - 2}}{b^2} + ....{C_n}{b^n} $ .
Substituting the value of a and b as 1 in the above equation, we get
$
\Rightarrow {\left( {a + b} \right)^n} = {C_0}{a^n} + {C_1}{a^{n - 1}}b + {C_2}{a^{n - 2}}{b^2} + ....{C_n}{b^n} \\
{\left( {1 + 1} \right)^n} = {C_0} \times {1^n} + {C_1} \times {1^{n - 1}} \times 1 + {C_2} \times {1^{n - 2}} \times {1^2} + ....{C_n} \times {1^n} \\
{2^n} = {C_0} + {C_1} + {C_2}..... + {C_n} - - - - (i) \\
$
According to the question, the sum of the coefficient is equals to 4096. So, substituting the value of the sum of the coefficient as 4096, we get
$
\Rightarrow {2^n} = {C_0} + {C_1} + {C_2}..... + {C_n} \\
\Rightarrow {2^n} = 4096 \\
\Rightarrow {2^n} = {2^{12}} \\
\Rightarrow n = 12 \\
$
As the value of n is 12 so, we can write
$ \Rightarrow {\left( {a + b} \right)^n} = {\left( {a + b} \right)^{12}} $
For the even raised power in the binomial expansion, the greatest coefficient is given as $ {}^n{C_{\left( {\dfrac{n}{2}} \right)}} $ .
So, substituting the value of greatest coefficient by substituting the value of n=12 in the function $ {}^n{C_{\left( {\dfrac{n}{2}} \right)}} $ .
$
\Rightarrow {C_g} = {}^n{C_{\left( {\dfrac{n}{2}} \right)}} \\
= {}^{12}{C_{\left( {\dfrac{{12}}{2}} \right)}} \\
= {}^{12}{C_6} \\
= \dfrac{{12!}}{{(12 - 6)!6!}} \\
= \dfrac{{12!}}{{6!6!}} \\
= \dfrac{{12 \times 11 \times 10 \times 9 \times 8 \times 7}}{{6 \times 5 \times 4 \times 3 \times 2}} \\
= 924 \\
$
Hence, the greatest coefficient is 924.
So, the correct answer is “Option B”.
Note: The total number of terms in the expansion of $ {\left( {x + y} \right)^n} $ is $ \left( {n + 1} \right) $ .
If the value of n is even, then the greatest coefficient term will be given as $ {}^n{C_{\left( {\dfrac{n}{2}} \right)}} $ .
If n is even, then the middle term is \[\left( {\dfrac{n}{2}} \right){\text{ and }}\left( {\dfrac{n}{2} + 1} \right)\]and if n is odd, then the middle term is \[\left( {\dfrac{{n + 1}}{2}} \right)\].
Complete step-by-step answer:
Binomial expansion theorem is a theorem which specifies the expansion of any power \[{\left( {a + b} \right)^n}\] of a binomial \[\left( {a + b} \right)\] as a sum of products e.g. \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]. The number of in the expansion depends upon the raised positive integral power. If the raised power of expansion is\[n\] then the number of terms of the expansion will be \[n + 1\].
The standard expansion of the Binomial theorem has been given as
$ {\left( {a + b} \right)^n} = {C_0}{a^n} + {C_1}{a^{n - 1}}b + {C_2}{a^{n - 2}}{b^2} + ....{C_n}{b^n} $ .
Substituting the value of a and b as 1 in the above equation, we get
$
\Rightarrow {\left( {a + b} \right)^n} = {C_0}{a^n} + {C_1}{a^{n - 1}}b + {C_2}{a^{n - 2}}{b^2} + ....{C_n}{b^n} \\
{\left( {1 + 1} \right)^n} = {C_0} \times {1^n} + {C_1} \times {1^{n - 1}} \times 1 + {C_2} \times {1^{n - 2}} \times {1^2} + ....{C_n} \times {1^n} \\
{2^n} = {C_0} + {C_1} + {C_2}..... + {C_n} - - - - (i) \\
$
According to the question, the sum of the coefficient is equals to 4096. So, substituting the value of the sum of the coefficient as 4096, we get
$
\Rightarrow {2^n} = {C_0} + {C_1} + {C_2}..... + {C_n} \\
\Rightarrow {2^n} = 4096 \\
\Rightarrow {2^n} = {2^{12}} \\
\Rightarrow n = 12 \\
$
As the value of n is 12 so, we can write
$ \Rightarrow {\left( {a + b} \right)^n} = {\left( {a + b} \right)^{12}} $
For the even raised power in the binomial expansion, the greatest coefficient is given as $ {}^n{C_{\left( {\dfrac{n}{2}} \right)}} $ .
So, substituting the value of greatest coefficient by substituting the value of n=12 in the function $ {}^n{C_{\left( {\dfrac{n}{2}} \right)}} $ .
$
\Rightarrow {C_g} = {}^n{C_{\left( {\dfrac{n}{2}} \right)}} \\
= {}^{12}{C_{\left( {\dfrac{{12}}{2}} \right)}} \\
= {}^{12}{C_6} \\
= \dfrac{{12!}}{{(12 - 6)!6!}} \\
= \dfrac{{12!}}{{6!6!}} \\
= \dfrac{{12 \times 11 \times 10 \times 9 \times 8 \times 7}}{{6 \times 5 \times 4 \times 3 \times 2}} \\
= 924 \\
$
Hence, the greatest coefficient is 924.
So, the correct answer is “Option B”.
Note: The total number of terms in the expansion of $ {\left( {x + y} \right)^n} $ is $ \left( {n + 1} \right) $ .
If the value of n is even, then the greatest coefficient term will be given as $ {}^n{C_{\left( {\dfrac{n}{2}} \right)}} $ .
If n is even, then the middle term is \[\left( {\dfrac{n}{2}} \right){\text{ and }}\left( {\dfrac{n}{2} + 1} \right)\]and if n is odd, then the middle term is \[\left( {\dfrac{{n + 1}}{2}} \right)\].
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