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The coefficient of $x^{12}$ in $\left(x^{3}+x^{4}+x^{5}+x^{6}+\dots \right)^{3}$ is ___________.
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(x^3+x^4+x^5+x^6+- - -)^3

(x^3(1+x+x^2+x^3+- - -))^3

x^9(1+x+x^2+x^3+ - - -)^3

[note : (1+x+x^2+x^3+ - - -)

 sum of infinite no: 1 / 1-r

r = x ]

==>x^9(1/(1-x)^3) = x^9((1-x)^-3)

==> x^9* 3-1+rCr            [ By n-1+rCr]

=> x^9*x^3(3-1+3C3)    [x^12 = x^9*x^3]

==> x^12(5C3)

==> 10x^12

Ans: 10
1 votes
1 votes

We have been given the sequence, $(x^3+x^4+x^5+x^6+...)^3$.

Start by taking $x^3$ common:

$(x^3+x^4+x^5+x^6+...)^3=\left [ x^3(1+x+x^2+x^3+...) \right ]^3$

This can now be written as follows by distributing the power:

$(x^3+x^4+x^5+x^6+...)^3=x^9*(1+x+x^2+x^3+...)^3$


Notice that:

$(1+x+x^2+x^3+...)^3 = \frac{1}{(1-x)^3}$


Thus, we now have:

$x^9*(1+x+x^2+x^3+...)^3=x^9*\frac{1}{(1-x)^3}$

Clearly, the coefficient of $x^{12}$ will be the coefficient of $x^3$ in the expansion of $\frac{1}{(1-x)^3}$. The goal now is to find this expansion.

Start with:

$\frac{1}{(1-x)} = 1+x+x^2+...+x^5+...$

Differentiate both sides with respect to $x$:

$\frac{1}{(1-x)^2} = 1+2x+...+5x^4+...$

Differentiate both sides with respect to $x$ againt:

$\frac{2}{(1-x)^3} = 2+...+20x^3+...$

Distribute the $2$ in numerator to the RHS:

$\frac{1}{(1-x)^3} = 1+...+10x^3+...$

Now, when you multiply this with $x^9$, the coeffiicient of $x^{12}$ will come out as $10$.

ANSWER: 10.

Answer:

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