Question

What is the coefficient of ${x^{10}}$ in the expansion of ${\left( {1 + x} \right)^2}{\left( {1 + {x^2}} \right)^3}{\left( {1 + {x^3}} \right)^4}$?
${\text{A}}{\text{. 52}} \\\ {\text{B}}{\text{. 44}} \\\ {\text{C}}{\text{. 50}} \\\ {\text{D}}{\text{. 56}} \\\$

Answer 1

Hint- Here, we will proceed by expanding ${\left( {1 + x} \right)^2}$, ${\left( {1 + {x^2}} \right)^3}$ and ${\left( {1 + {x^3}} \right)^4}$ individually using the binomial theorem and then, we will multiply all these terms together and consider only the terms which will contain ${x^{10}}$ in the final expansion.

Complete step by step answer:
According to binomial theorem,
${\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1}\left( x \right) + {}^n{C_2}\left( {{x^2}} \right) + ... + {}^n{C_{n - 1}}\left( {{x^{n - 1}}} \right) + {}^n{C_n}\left( {{x^n}} \right){\text{ }} \to (1{\text{)}}$ where ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
By putting n = 2 in the formula given by equation (1), we have
${\left( {1 + x} \right)^2} = {}^2{C_0} + {}^2{C_1}\left( x \right) + {}^2{C_2}\left( {{x^2}} \right){\text{ }} \to {\text{(2)}}$
By putting n = 3 and replacing x with ${x^2}$ in the formula given by equation (1), we have

$${\left( {1 + {x^2}} \right)^3} = {}^3{C_0} + {}^3{C_1}\left( {{x^2}} \right) + {}^3{C_2}\left[ {{{\left( {{x^2}} \right)}^2}} \right] + {}^3{C_3}\left[ {{{\left( {{x^2}} \right)}^3}} \right] \\\ \Rightarrow {\left( {1 + {x^2}} \right)^3} = {}^3{C_0} + {}^3{C_1}\left( {{x^2}} \right) + {}^3{C_2}\left( {{x^4}} \right) + {}^3{C_3}\left( {{x^6}} \right){\text{ }} \to {\text{(3)}} \\\$$

By putting n = 4 and replacing x with ${x^3}$ in the formula given by equation (1), we have

$${\left( {1 + {x^3}} \right)^4} = {}^4{C_0} + {}^4{C_1}\left( {{x^3}} \right) + {}^4{C_2}\left[ {{{\left( {{x^3}} \right)}^2}} \right] + {}^4{C_3}\left[ {{{\left( {{x^3}} \right)}^3}} \right] + {}^4{C_4}\left[ {{{\left( {{x^3}} \right)}^4}} \right] \\\ \Rightarrow {\left( {1 + {x^3}} \right)^4} = {}^4{C_0} + {}^4{C_1}\left( {{x^3}} \right) + {}^4{C_2}\left( {{x^6}} \right) + {}^4{C_3}\left( {{x^9}} \right) + {}^4{C_4}\left( {{x^{12}}} \right){\text{ }} \to {\text{(4)}} \\\$$

Multiplying all three equations (2), (3) and (4) together, we get

$$\Rightarrow {\left( {1 + x} \right)^2}{\left( {1 + {x^2}} \right)^3}{\left( {1 + {x^3}} \right)^4} = \left( {{}^2{C_0} + {}^2{C_1}x + {}^2{C_2}{x^2}} \right)\left( {{}^3{C_0} + {}^3{C_1}{x^2} + {}^3{C_2}{x^4} + {}^3{C_3}{x^6}} \right)\left( {{}^4{C_0} + {}^4{C_1}{x^3} + {}^4{C_2}{x^6} + {}^4{C_3}{x^9} + {}^4{C_4}{x^{12}}} \right) \\\ \Rightarrow {\left( {1 + x} \right)^2}{\left( {1 + {x^2}} \right)^3}{\left( {1 + {x^3}} \right)^4} = \left( {1 + 2x + {x^2}} \right)\left( {1 + 3{x^2} + 3{x^4} + {x^6}} \right)\left( {1 + 4{x^3} + 6{x^6} + 4{x^9} + {x^{12}}} \right) \\\ \Rightarrow {\left( {1 + x} \right)^2}{\left( {1 + {x^2}} \right)^3}{\left( {1 + {x^3}} \right)^4} = \left( {1 + 2x + {x^2} + 3{x^2} + 6{x^3} + 3{x^4} + 3{x^4} + 6{x^5} + 3{x^6} + {x^6} + 2{x^7} + {x^8}} \right)\left( {1 + 4{x^3} + 6{x^6} + 4{x^9} + {x^{12}}} \right) \\\ \Rightarrow {\left( {1 + x} \right)^2}{\left( {1 + {x^2}} \right)^3}{\left( {1 + {x^3}} \right)^4} = \left( {1 + 2x + 4{x^2} + 6{x^3} + 6{x^4} + 6{x^5} + 4{x^6} + 2{x^7} + {x^8}} \right)\left( {1 + 4{x^3} + 6{x^6} + 4{x^9} + {x^{12}}} \right) \\\$$

Now we want the coefficient of ${x^{10}}$ so in above equation $\left( {2x} \right)$ multiplied by $$$$ similarly,
$\left( {6{x^4}} \right)$ multiplied by $\left( {6{x^6}} \right)$ and $\left( {2{x^7}} \right)$ multiplied by $\left( {4{x^3}} \right)$ to get the term ${x^{10}}$.
So the coefficient of ${x^{10}}$ $= \left( {2 \times 4} \right) + \left( {6 \times 6} \right) + \left( {2 \times 4} \right)$
$\Rightarrow$Coefficient of ${x^{10}}$ in the expansion of ${\left( {1 + x} \right)^2}{\left( {1 + {x^2}} \right)^3}{\left( {1 + {x^3}} \right)^4} = 8 + 36 + 8 = 52$

Hence option (A) is correct.

Note- In this particular problem, in the final expansion i.e., ${\left( {1 + x} \right)^2}{\left( {1 + {x^2}} \right)^3}{\left( {1 + {x^3}} \right)^4} = \left( {1 + 2x + 4{x^2} + 6{x^3} + 6{x^4} + 6{x^5} + 4{x^6} + 2{x^7} + {x^8}} \right)\left( {1 + 4{x^3} + 6{x^6} + 4{x^9} + {x^{12}}} \right)$, only those terms are multiplied which are giving ${x^{10}}$. Then, all the coefficients are algebraically added (i.e., along with the sign) in order to get the final coefficient of ${x^{10}}$ because here ${x^{10}}$ would be taken outside since, it is common to all these terms.