Question

Coefficient of ${x^{11}}$ in the expansion of ${\left( {1 + {x^2}} \right)^4}{\left( {1 + {x^3}} \right)^7}{\left( {1 + {x^4}} \right)^{12}}$ is
A. 1051
B. 1106
C. 1113
D. 1120

Answer 1

Binomial Theorem is used to expand binomial expression that has been raised to some higher power.
Use the general term of binomial expansion to find the coefficient of particular power.
Remember that while multiplying two terms their powers get added only when their bases are the same.
For example: $\left( {2{x^3}} \right)\left( {4{x^4}} \right)$
$\Rightarrow \left( {2 \times 4} \right){x^{3 + 4}} \\\ \Rightarrow 8{x^7} \\\$
We know, combination: $C(n,r) = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Let’s understand the term ‘coefficient’ by an example:
Example: Find the coefficient of ${x^2}$ in the expression $2{x^3} + 4{x^2}y - x$
The term including ${x^2}$ is $4{x^2}y$
The coefficient of ${x^2}$ is $4y$
Thus, the coefficient of ${x^2}$ is the remaining part of the term except ${x^2}$.

Complete step-by-step answer:
The binomial expansion:
${\left( {a + b} \right)^n} = C\left( {n,0} \right){a^n} + C\left( {n,1} \right){a^{n - 1}}b + C\left( {n,2} \right){a^{n - 2}}{b^2} + ... + C(n,n){b^n}$
Where ‘a’ is the first term;
‘b’ is the second term;
‘n’ is index;
‘C(n,r)” is the number of ways in which we can choose r objects out of a set containing n different objects such that the order of selection does not matter.
The general term ${T_{k + 1}}$ for the ${\left( {k + 1} \right)^{th}}$term of the binomial expansion is:
${T_{k + 1}} = C\left( {n,k} \right){a^{n - k}}{b^k}$
Thus, the binomial expansion can also be written as
${\left( {a + b} \right)^n} = \sum\limits_{k = 0}^n {C\left( {n,k} \right){a^{n - k}}{b^k}}$

The general term of: ${\left( {1 + {x^2}} \right)^4}$
${T_{a + 1}} = C\left( {4,a} \right){1^{4 - a}}{\left( {{x^2}} \right)^a} \\\ \Rightarrow C\left( {4,a} \right){x^{2a}} \\\$
Binomial expansion of ${\left( {1 + {x^2}} \right)^4}$

$${\left( {1 + {x^2}} \right)^4} = \sum\limits_{a = 0}^4 {C\left( {4,a} \right){1^{4 - a}}{{\left( {{x^2}} \right)}^a}} \\\ \Rightarrow \sum\limits_{a = 0}^4 {C\left( {4,a} \right){x^{2a}}} \\\$$

General term of ${\left( {1 + {x^3}} \right)^7}$
${T_{b + 1}} = C\left( {7,b} \right){1^{7 - b}}{\left( {{x^3}} \right)^b} \\\ \Rightarrow C\left( {7,b} \right){x^{3b}} \\\$
Binomial expansion of ${\left( {1 + {x^3}} \right)^7}$

$${\left( {1 + {x^3}} \right)^7} = \sum\limits_{b = 0}^7 {C\left( {7,b} \right){1^{7 - b}}{{\left( {{x^3}} \right)}^b}} \\\ \Rightarrow \sum\limits_{b = 0}^7 {C\left( {7,b} \right){x^{3b}}} \\\$$

General term of ${\left( {1 + {x^4}} \right)^{12}}$
${T_{c + 1}} = C\left( {12,c} \right){1^{12 - c}}{\left( {{x^4}} \right)^c} \\\ \Rightarrow C\left( {12,c} \right){x^{4c}} \\\$
Binomial expansion of ${\left( {1 + {x^4}} \right)^{12}}$

$${\left( {1 + {x^4}} \right)^{12}} = \sum\limits_{c = 0}^{12} {C\left( {12,c} \right){1^{12 - c}}{{\left( {{x^4}} \right)}^c}} \\\ \Rightarrow \sum\limits_{c = 0}^{12} {C\left( {12,c} \right){x^{4c}}} \\\$$

The expansion of ${\left( {1 + {x^2}} \right)^4}{\left( {1 + {x^3}} \right)^7}{\left( {1 + {x^4}} \right)^{12}}$
$\Rightarrow \left[ {\sum\limits_{a = 0}^4 {C\left( {4,a} \right){x^{2a}}} } \right]\left[ {\sum\limits_{b = 0}^7 {C\left( {7,b} \right){x^{3b}}} } \right]\left[ {\sum\limits_{c = 0}^{12} {C\left( {12,c} \right){x^{4c}}} } \right]$

Step 2: Find the coefficient of ${x^{11}}$
${x^{2a}} \times {x^{3b}} \times {x^{4c}} = {x^{11}}$
${x^{2a + 3b + 4c}} = {x^{11}}$
$\Rightarrow 2a + 3b + 4c = 11$
Coefficient of ${x^{2a}}$ $= C\left( {4,a} \right)$
Coefficient of ${x^{3b}}$ $= C\left( {7,b} \right)$
Coefficient of ${x^{4c}}$ $= C\left( {12,c} \right)$
Thus, the coefficient of ${x^{2a + 3b + 4c}}$ $= C\left( {4,a} \right) \times C\left( {7,b} \right) \times C\left( {12,c} \right)$

Choose the values of a, b, and c such that the sum of 2a, 3b, and 4c is 11
$11 = 1 + 10 \\\ \Rightarrow 2 + 9 \\\ \Rightarrow 3 + 8 \\\ \Rightarrow 4 + 7 \\\ \Rightarrow 5 + 6 \\\$

For a = 1, b = 3, c = 0
$\Rightarrow 2\left( 1 \right) + 3\left( 3 \right) + 4\left( 0 \right) = 11$
Thus, the coefficient of ${x^{2\left( 1 \right) + 3\left( 3 \right) + 4\left( 0 \right)}},i.e.{\text{ }}{x^{11}}$ $= C\left( {4,1} \right) \times C\left( {7,3} \right) \times C\left( {12,0} \right)$
$\Rightarrow \dfrac{{4!}}{{1!\left( 3 \right)!}} \times \dfrac{{7!}}{{3!\left( 4 \right)!}} \times \dfrac{{12!}}{{0!\left( {12} \right)!}} \\\ \Rightarrow \dfrac{{4 \times 3!}}{{3!}} \times \dfrac{{7 \times 6 \times 5 \times 4!}}{{3 \times 2 \times 1 \times 4!}} \times \dfrac{{12!}}{{1\left( {12} \right)!}} \\\ \Rightarrow 4 \times 35 \times 1 \\\ \Rightarrow 140 \\\$

For a = 2, b = 1, c = 1
$\Rightarrow 2\left( 2 \right) + 3\left( 1 \right) + 4\left( 1 \right) \\\ = 4 + 3 + 4 = 11 \\\$
Thus, the coefficient of ${x^{2\left( 2 \right) + 3\left( 1 \right) + 4\left( 1 \right)}},i.e.{\text{ }}{x^{11}}$ $= C\left( {4,2} \right) \times C\left( {7,1} \right) \times C\left( {12,1} \right)$
$\Rightarrow \dfrac{{4!}}{{2!\left( 2 \right)!}} \times \dfrac{{7!}}{{1!\left( 6 \right)!}} \times \dfrac{{12!}}{{1!\left( {11} \right)!}} \\\ \Rightarrow \dfrac{{4 \times 3 \times 2!}}{{2 \times 1 \times 2!}} \times \dfrac{{7 \times 6!}}{{6!}} \times \dfrac{{12 \times 11!}}{{11!}} \\\ \Rightarrow 6 \times 7 \times 12 \\\ \Rightarrow 504 \\\$

For a = 4, b = 1, c = 0
$\Rightarrow 2\left( 4 \right) + 3\left( 1 \right) + 4\left( 0 \right) \\\ = 8 + 3 + 0 = 11 \\\$
Thus, the coefficient of ${x^{2\left( 4 \right) + 3\left( 1 \right) + 4\left( 0 \right)}},i.e.{\text{ }}{x^{11}}$ $= C\left( {4,4} \right) \times C\left( {7,1} \right) \times C\left( {12,0} \right)$

$$\Rightarrow \dfrac{{4!}}{{4!\left( 0 \right)!}} \times \dfrac{{7!}}{{1!\left( 6 \right)!}} \times \dfrac{{12!}}{{0!\left( {12} \right)!}} \\\ \Rightarrow \dfrac{{4!}}{{4!}} \times \dfrac{{7 \times 6!}}{{6!}} \times \dfrac{{12!}}{{\left( {12} \right)!}} \\\ \Rightarrow 1 \times 7 \times 1 \\\ \Rightarrow 7 \\\$$

For a = 0, b = 1, c = 2
$\Rightarrow 2\left( 0 \right) + 3\left( 1 \right) + 4\left( 2 \right) \\\ = 0 + 3 + 8 = 11 \\\$
Thus, the coefficient of ${x^{2\left( 0 \right) + 3\left( 1 \right) + 4\left( 2 \right)}},i.e.{\text{ }}{x^{11}}$ $= C\left( {4,0} \right) \times C\left( {7,1} \right) \times C\left( {12,2} \right)$

$$\Rightarrow \dfrac{{4!}}{{0!\left( 4 \right)!}} \times \dfrac{{7!}}{{1!\left( 6 \right)!}} \times \dfrac{{12!}}{{2!\left( {10} \right)!}} \\\ \Rightarrow \dfrac{{4!}}{{4!}} \times \dfrac{{7 \times 6!}}{{6!}} \times \dfrac{{12 \times 11 \times 10!}}{{2 \times 1 \times 10!}} \\\ \Rightarrow 1 \times 7 \times 66 \\\ \Rightarrow 462 \\\$$

Thus, the coefficient of ${x^{11}}$ in the expansion of ${\left( {1 + {x^2}} \right)^4}{\left( {1 + {x^3}} \right)^7}{\left( {1 + {x^4}} \right)^{12}}$ is the sum of the coefficient of ${x^{2\left( 1 \right) + 3\left( 3 \right) + 4\left( 0 \right)}}$ and coefficient of ${x^{2\left( 2 \right) + 3\left( 1 \right) + 4\left( 1 \right)}}$ and coefficient of ${x^{2\left( 4 \right) + 3\left( 1 \right) + 4\left( 0 \right)}}$ and coefficient of ${x^{2\left( 0 \right) + 3\left( 1 \right) + 4\left( 2 \right)}}$ .
the coefficient of ${x^{11}} = 140 + 504 + 7 + 462$
$= 1113$
The coefficient of ${x^{11}}$in the expansion of ${\left( {1 + {x^2}} \right)^4}{\left( {1 + {x^3}} \right)^7}{\left( {1 + {x^4}} \right)^{12}}$ is 1113.

So, the correct answer is “Option C”.

Note: Index, n of the binomial expression must be a positive integer.
The total number of terms in a binomial expression of index n is n + 1.
Students often choose the same value of r in the expansion instead of different like a, b, and c in the above solution. Then the expression will be
${x^{2r}} \times {x^{3r}} \times {x^{4r}} = {x^{11}}$
${x^{7r}} = {x^{11}}$
$7r = 11 \\\ \Rightarrow r = \dfrac{{11}}{7} \\\$
This is a wrong step.
The expression ${\left( {1 + {x^2}} \right)^4}{\left( {1 + {x^3}} \right)^7}{\left( {1 + {x^4}} \right)^{12}}$ has the expansion of three different expressions ${\left( {1 + {x^2}} \right)^4}$, ${\left( {1 + {x^3}} \right)^7}$ , and ${\left( {1 + {x^4}} \right)^{12}}$.