Let $a_k$ be the coefficient of $x^k$ in the expansion of :- $(x+1) + (x+1)^2 + (x+1)^3 + (x+1)^4 + ... + (x+1)^{99}$ . Find $[\frac{a_4}{a_3}]$

Question

Let $a_k$ be the coefficient of $x^k$ in the expansion of :- $(x+1) + (x+1)^2 + (x+1)^3 + (x+1)^4 + ... + (x+1)^{99}$ . Find $[\frac{a_4}{a_3}].$

What I Tried :- Honestly there's nothing I can do to start here . The problem is that I don't know how to find the coefficient of any term of any expression (assuming the term is there in the expression) . Once I know that type of method, this problem will be straight-forward for me .

So can someone say me some type of method ?

Answer 1

You can use the formula for the sum of a geometric progression and then the binomial theorem to find that your sum is $$ \frac{{1 - (x + 1)^{100} }}{{1 - (x + 1)}} - 1 = \frac{{(x + 1)^{100} - 1}}{x} - 1 = \left( {\sum\limits_{k = 1}^{100} {\binom{100}{k}x^{k - 1} } } \right) - 1 \\ = 99 + \sum\limits_{k = 2}^{100} {\binom{100}{k}x^{k - 1} } = 99 + \sum\limits_{k = 1}^{99} {\binom{100}{k+1}x^{k} }. $$ Thus $$ \frac{{a_4 }}{{a_3 }} = \frac{\binom{100}{5}}{\binom{100}{4}} = \frac{{96}}{5} = 19.2. $$