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

  1. $2^{14}$
  2. $31$
  3. $\left ( \frac{3}{3} \right ) + \left ( \frac{10}{1} \right )$
  4. $\left ( \frac{3}{3} \right ) + 2\left ( \frac{10}{1} \right )$
  5. $\left ( \frac{3}{3} \right ) \left ( \frac{10}{1} \right ) 2^{9}$
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29 votes

Using the Binomial Theorem:

${{(x+y)^{n}=\sum_{r=0}^{n}\binom{n}{r}x^{n-r}y^{r}} }$

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

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

Case$1:$If $r=3 , k=0$ we get $x^{3}$

So, Coefficient of $x^{3}$ is =$\binom{3}{3}$$\cdot \binom{10}{0}(2^{10})$ = $2^{10}$

Case$2:$If $r=1 , k=1 $ we get $x^{3}$

So, Coefficient of $x^{3}$ is = $\binom{3}{1}\cdot \binom{10}{1}(2)^{9}$ = $30\times 2^{9}$

So, finally Coefficient of  $x^{3}$ is $ = 2^{10} + 30 \times 2^{9}$

$\qquad \qquad = 2^{9}( 2 + 30 )$

$\qquad \qquad = 2^{9}\times 32$

$\qquad \qquad=2^{9}\times 2^{5}$

$\qquad \qquad=2^{14}$

Correct Answer: A

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$\;\;(1+x)^3=(1+x^3+3x+3x^2)$

and

$(2+x^2)^{10} = ^{10}C_0*2^0*(x^2)^{10}+^{10}C_1*2^1*(x^2)^9+\ldots+^{10}C_9*2^9*(x^2)^1+ ^{10}C_{10}*2^{10}*(x^2)^0$

$\text{So,coefficient of }x^3=^{10}C_{10}*2^{10}+3*^{10}C_9*2^9$

$=2^9(32)=2^{14}.$

As here we need to multiply last term of second expansion with first term of first coefficient $x^3$ and $3x$ with $x^2$ in the second expansion.
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it should be 2^14

equivallent exp:(1+x)3210(1+x2/2)10 

 we can get x^3 in the expansion if (1+x),(1+x) and (1+x)  is multiplied together or one of the (1+x)  block multiplied with one of the (2+x^2)  block

so,coff of x^3= (3c3 + 3c1*10c1*1/2).*2^10  =  (1+3*10*1/2)*2^10  =2^14

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(1+x)3 (2+ x2 )10

Lets Expand this ->

(1+x)(1+x) (1+x) (2+ x2 )(2+ x2 )(2+ x2 )(2+ x2 )(2+ x2 )(2+ x2 )(2+ x2 )(2+ x2 )(2+ x2 )(2+ x2 )

If you see here there are two ways of getting x3

Way 0 :->Choose x from each of (1+x)3 In that case we need to choose 2 from each of (2+ x2 ). Here we get 210 part of sum.

Way 1-> Here we can choose (1+x) from (1+x)3 . Which we can do in 3 ways (3 choose 1) . Then we need to choose 1 xfrom (2+ x2 )10 , Which we can do in 10 way ( 10 choose 1)

Total ways we can choose x3 using way 1 is 30 =(10* 3)

Total sum of coeffient in way 1 = 29 * 30

After calculation we get = 210 + 29 * 30 => 16384 => 214 => Answer A)

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