سوال ۳۲ کنکور ارشد علوم کامپیوتر ۱۴۰۰

The radius of convergence $R$ of $\sum \alpha_n (x+1)^n$ depends only on the coefficients $\alpha_n$ (centering at $x=-1$ does not change $R$). By the ratio form of Cauchy–Hadamard,

$\frac{1}{R}=\lim_{n\to\infty}\left|\frac{\alpha_{n+1}}{\alpha_n}\right|\quad(\text{when this limit exists}).$

Step 1 — close the inner sum. Factor out the part of $2^{\,n-k}$ that does not depend on $k$:

$\alpha_n=\sum_{k=0}^{n}2^{\,n-k}\bigl(3k^{2}+k-1\bigr)=2^{n}\sum_{k=0}^{n}\frac{3k^{2}+k-1}{2^{k}}.$

As $n\to\infty$ the inner sum converges, because $\displaystyle\sum_{k=0}^{\infty}\frac{3k^{2}+k-1}{2^{k}}$ is a convergent series (polynomial $\times$ geometric with ratio $\tfrac12<1$). Its value is finite — call it $C$. Using $\sum k^2/2^k=6$, $\sum k/2^k=2$, $\sum 1/2^k=2$:

$C=3(6)+1(2)-1(2)=18.$

So in closed form $\alpha_n=18\cdot 2^{n}-3n^{2}-13n-19$, and the dominant term is $18\cdot 2^{n}$:

$\alpha_n\sim 18\cdot 2^{n}\quad(n\to\infty).$

Step 2 — take the ratio.

$\frac{\alpha_{n+1}}{\alpha_n}\;\longrightarrow\;\frac{18\cdot 2^{n+1}}{18\cdot 2^{n}}=2.$

(Equivalently $\lim_{n\to\infty}|\alpha_n|^{1/n}=2$ since the geometric factor $2^{n}$ controls the growth.)

Step 3 — invert.

$R=\frac{1}{\,\lim |\alpha_{n+1}/\alpha_n|\,}=\frac{1}{2}.$

The answer is option 4.

Exam trick: A polynomial-in-$k$ times $r^{\,n-k}$ summed to $n$ behaves like $C\cdot r^{n}$ for large $n$ (the tail of the geometric series is a constant). Only the geometric base $r$ matters for the radius, so $R=1/r=1/2$ here — the polynomial $3k^2+k-1$ is irrelevant to the radius.