Let S be the sum of the first 9 terms of the series x+ka+x^2+(k+2)a+x^3+(k+4)a+x^4+(k+6)a+…

Question

Let $S$ be the sum of the first 9 terms of the series $\{x+ka\}+\{x^2+(k+2)a\}+\{x^3+(k+4)a\}+\{x^4+(k+6)a\}+\cdots$, where $a
e0$ and $x
e1$. If $S = \dfrac{x^{10}-x+45a(x-1)}{x-1}$, then $k$ is equal to:

Accepted Answer

$$S = (x+x^2+\cdots+x^9) + a\big[k+(k+2)+\cdots\text{ (9 terms)}\big] = \frac{x^{10}-x}{x-1}+\frac{9a\big(2k+8\times2\big)}{2}.$$

$$= \frac{x^{10}-x+9a(k+8)(x-1)}{x-1}.$$

Comparing with the given form: $9a(k+8)=45a \Rightarrow k+8=5 \Rightarrow k=-3.$

Solutions · 1