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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. $\left ( \frac{3}{3} \right ) \left ( \frac{10}{1} \right ) 2^{9}$

edited ago | 345 views

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

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.
answered by Boss (5.7k points)
edited ago

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

answered by Active (2.2k points)

(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)

answered by Veteran (46.8k points)
edited

Nice solution by using Combinatorics concept.

Just small correction in this line "Total ways we can choose x3 using way 1 is 30 ! (10* 3)" .

It should be 30 not 30! .