What is coefficient of $x^{17}$ in expansion of $(1+x)^{100}+ x(1+x)^{99}+ x^2(1+x)^{98}+ ... +x^{100}$?

Question

Please answer this question. In this question, at the place of $(1+2x)^{100}$ this whole thing was given $$ (1+x)^{100} + x \cdot (1+x)^{99}+ ... + x^{100} $$ and at place of this I wrote $(1+2x)^{100}$. Sorry I realised that I did it wrong so I changed my question. Please answer it now.

Answer 1

Setting $1+x=y$ the expression can be written as

$y^{100}+xy^{99}+\ldots+x^{100}$

which is

$$\frac{y^{101}-x^{101}}{y-x}=y^{101}-x^{101}=(1+x)^{101}-x^{101}$$ and the coefficient of $x^{17}$ is $\binom{101}{17}$

Answer 2

$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k}a^kb^{n-k}$

So in our case...

$(1+2x)^{100} = \sum_{k=0}^{100}\binom{100}{0}1^k2^{n-k}x^{n-k} = \sum_{k=0}^{100}\binom{100}{k}2^{100-k}x^{100-k}$

We are interested in the coefficent of $x^{17}$, that happens when $k=83$.

So the answer should be $\binom{100}{83}2^{17}$

Answer 3

$$(1+2x)=\sum_{k=0}^{100}\binom{100}{k}(2x)^k=\sum_{k=0}^{100}\left[\binom{100}{k}2^k\right]x^k$$