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What is the coefficient of $\large\color{green}{x^{6}}$ in the following series expansion?

$$\color{maroon}{\begin{align*} \frac{1}{1-x}.\frac{1}{1-x^2}.\frac{1}{1-x^3}........ \end{align*}}$$
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$\large \color{navy}{\frac{1}{1-x}= 1 +x + x^2 + x^3 + x^4 \dots + \infty }$
Simplifying each term, we get an equivalent expression for finding coefficient of $x^6$. Finding coefficient of $x^6$ in $\color{green}{\frac{1}{1-x}\frac{1}{1-x^2}\frac{1}{1-x^3}}\dots$ is equivalent to finding coefficient of $x^6$ in :

$\color{navy}{\begin{align*}(1+x+x^2 + \dots + x^6)(1+x^2 +x^4+x^6)(1+x^3+x^6)(1+x^4)(1+x^5)(1+x^6)\end{align*}}$

We are interested in coefficient of $x^6$. So, neglect higher powers.

$\Rightarrow (1+x+x^2 + \dots + x^6)(1+x^2 +x^4+x^6)(1+x^3+x^6)(1 + x^4 + x^5 + x^6)$

$\Rightarrow (1+x+x^2 + \dots + x^6)(1+x^2 +x^4+x^6)(1 + x^3 + x^4 + x^5 + 2x^6)$

$\Rightarrow (1+x+x^2 + x^3 + x^4 + x^5 + x^6)(1+x^2 + x^3 + 2x^4 + 2x^5 + 4x^6)$

Coefficient of $x^6$ $=  4+2+2+1+1+1 = 11$
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