TIFR CSE 2010 | Part A | Question: 12

Question

The coefficient of $x^{3}$ in the expansion of $(1 + x)^{3} (2 + x^{2})^{10}$ is.

• A. $2^{14}$
• B. $31$
• C. $\left ( \frac{3}{3} \right ) + \left ( \frac{10}{1} \right )$
• D. $\left ( \frac{3}{3} \right ) + 2\left ( \frac{10}{1} \right )$
• E. $\left ( \frac{3}{3} \right ) \left ( \frac{10}{1} \right ) 2^{9}$

Comments

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Answer 1

Just two cases Possible here :

1) Extract x from (1+X)³ and x² from (2+x²)¹⁰ to finally get x³.  --> ³C₁ * ¹⁰C₁ * 2⁹

2) Extract x³ from (1+X)³ and x⁰ from (2+x²)¹⁰ to finally get x³.  --> ³C₃ * ¹⁰C₀ * 2¹⁰

So coefficient of x³ is --> case 1 + case 2 --> 30*2⁹ + 2¹⁰ --> 32 * 2⁹ --> 2¹⁴.

Answer 2

Using Binomial Theorem:

$(a+b)^{n} = \sum_{k=0}^{n}\binom{n}{k} (a)^{n-k}.(b)^{k}$

$(1+x)^{3}.(2+x^{2})^{10} = \sum_{k=0}^{3}\binom{3}{k}(1)^{3-k}.(x)^{k}$  $\ast$ $\sum_{r=0}^{10}\binom{10}{r}(2)^{10-r}.(x^{2})^{r}$

$(1+x)^{3}.(2+x^{2})^{10} = \sum_{k=0}^{3}\binom{3}{k}(x)^{k}$  $\ast$ $\sum_{r=0}^{10}\binom{10}{r}(2)^{10-r}.(x^{2})^{r}$

Case$(1):$ Put $k=3$ and $r=0$,we get $x^{3}$

Coefficient of $x^{3}:\binom{3}{3}\ast\binom{10}{0}.(2)^{10} = 2^{10}$

Case$(2):$ Put $k=1$ and $r=1$,we get $x^{3}$

Coefficient of $x^{3}:\binom{3}{1}\ast\binom{10}{1}.(2)^{9} = 3\cdot10\cdot2^{9} = 30\cdot2^{9}$

Finally Coefficient of $x^{3}: 2^{10} + 30\cdot2^{9}$

                                    $\Rightarrow 2^{9}(2^{1} + 30)$

                                   $\Rightarrow 2^{9}(2 + 30)$

                                  $\Rightarrow 2^{9}(32)$

                                  $\Rightarrow 2^{9}(2^{5})$

                                 $\Rightarrow 2^{(9 + 5)}$

                                  $\Rightarrow 2^{14}$

  So ,correct answer is  $ 2^{14}$

 

Answer 3

Best way to do this

Using simple mathematics