A sum related to binomial theorem

Question

If $\dfrac{x^2+x+1}{1-x} = a_0+a_1x+a_2x^2+\cdots$

then $\displaystyle\sum_{\gamma = 1}^{50}a_{\gamma} = ??$

Original Image

This is a sum related to evaluating a series, from the chapter binomial theorem.

This is how I tried to solve the question

I took the $(1-x)$ from the denominator to the numerator making it a $$(1-x)^{-1}$$ and then expanded it binomially, and I tried to find the sum of the coefficients of $$x,x^2,x^3,....,x^{50}$$ However my answer isn't matching. I got $1325$ but the answer is $149$.

update:i do get my mistake, but how do i go about solving it?

Answer 1

Your first few terms are correct. But you only have three terms in the numerator, so the series is $1+2x+3x^2+3x^4+3x^5+...$

Answer 2

The general expansion is $(1-x)^{-1}=1+x+x^2...\infty$ so now we need sum of coefficients upto $50$. If we see the pattern we see that $(x^2+x+2)(1)=3$ sum of coeeficients. For $(1+x)(1+x+x^2)$ its 6 so its an AP. So we need sum upto $50$ which is $49\times 3=147$ but 1 term will be $x^{51}$ which we dont need so we get only $+2$ for last term so its $147+2=149$ . Hope its clear.

Answer 3

By considering,

$\begin{array}{rcccccc} \frac{1}{1-x} & = & 1 & + & x & + & x^{2}+\ldots + x^{50} + \ldots \\ \frac{x}{1-x} & = & & & x & + & x^{2}+\ldots + x^{50} + \ldots \\ \frac{x^2}{1-x} & = & & & & & x^{2}+\ldots + x^{50} + \ldots \\ \end{array}$

Hence the sum is $50+50+49=149$.