Question

find the coefficient of ${{x}^{9}}$ in $(1+x)(1+{{x}^{2}})(1+{{x}^{3}})....(1+{{x}^{100}})$

Answer 1

Now in this question first we will note the possible combinations by which ${{x}^{9}}$ is formed.
Now once we have all possible combinations we will add the coefficients of all combinations to find the coefficient of ${{x}^{9}}$ in $(1+x)(1+{{x}^{2}})(1+{{x}^{3}})....(1+{{x}^{100}})$

Complete step by step answer:
Now first Let us understand how this binomial opens up
First of all we know that ${{x}^{m}}{{x}^{n}}={{x}^{m+n}}$ .
Now in the expansion of $(1+x)(1+{{x}^{2}})(1+{{x}^{3}})....(1+{{x}^{100}})$ each power of x will multiply to 1 and other powers of x.
Now we need to find out how many ways ${{x}^{9}}$ can be formed
For understanding take example ${{x}^{3}}{{x}^{6}}={{x}^{9}}$. So here to make ${{x}^{9}}$ we would need ${{x}^{3}}$ and ${{x}^{6}}$ but this is just one combination.
Now the coefficient of ${{x}^{9}}$ will be sum of all the coefficients in which ${{x}^{9}}$ can be formed
We will find all such combinations now
So let us start with combinations made of two terms
Now the possibilities are {(1, 8), (2, 7), (3, 6), (4, 5)} since we have 1 + 8 = 2 + 7 = 3 + 6 = 4 + 5 = 9
That is our terms will be ${{x}^{1}}{{x}^{8}},{{x}^{2}}{{x}^{7}},{{x}^{3}}{{x}^{6}},{{x}^{4}}{{x}^{5}}$ now note that all this terms will have coefficient one since the only constant that any power of x could be multiplied in expansion of $(1+x)(1+{{x}^{2}})(1+{{x}^{3}})....(1+{{x}^{100}})$ is 1.
Now the coefficient here will be
$\begin{aligned} & {{x}^{1}}{{x}^{8}}+{{x}^{2}}{{x}^{7}}+{{x}^{3}}{{x}^{6}}+{{x}^{4}}{{x}^{5}} \\\ & =(1+1+1+1){{x}^{9}} \\\ & =4{{x}^{9}}......................(1) \\\ \end{aligned}$
But now we have just consider two terms together, there can also be cases where three terms come together for example ${{x}^{1}}{{x}^{2}}{{x}^{6}}$.
Now let us find all such combinations.
{(1, 2, 6), (1, 3, 5), (2, 3, 4)}
Hence we have three such combinations i.e. ${{x}^{1}}{{x}^{2}}{{x}^{6}},{{x}^{1}}{{x}^{3}}{{x}^{5}},{{x}^{2}}{{x}^{3}}{{x}^{4}}$
Now the coefficient of these terms is also 1 hence the final coefficient that we get will be
$\begin{aligned} & (1+1+1){{x}^{9}} \\\ & =3{{x}^{9}}..............(2) \\\ \end{aligned}$
Also note that there will also be a single term ${{x}^{9}}$
Hence we also have $1{{x}^{9}}...............(3)$
Hence from equation (1), equation (2) and equation (3) we get the final coefficient of ${{x}^{9}}$ is 8

So, the correct answer is “8”.

Note: Here while taking cases we got 4 cases while taking combination of two terms and 3 cases while taking combination of 3 terms. Now this is because we have not considered repeating combinations for example (1, 2) and (2, 1) means ${{x}^{2}}{{x}^{8}}$ and ${{x}^{8}}{{x}^{2}}$ . which are same thing hence we do not take same combinations. Similarly we have skipped repeating combinations in the cases where we took combinations of three terms together. Now also note that we have skipped the terms (2, 2, 4) or (4, 4, 1) and etc. this is because we cant have ${{x}^{4}}$ or ${{x}^{2}}$ two times in the expansion of $(1+x)(1+{{x}^{2}})(1+{{x}^{3}})....(1+{{x}^{100}})$ since we just have all unique powers here.