سوال ۱۱۲ کنکور ارشد علوم کامپیوتر ۱۴۰۲

Recognize the binomial theorem

The sum has exactly the shape of the binomial expansion of $(1+x)^{n}$:

$\sum_{i=0}^{n}\binom{n}{i}x^{i} = (1+x)^{n}.$

Here $n = 10$ and $x = 2$, so

$A = \sum_{i=0}^{10}\binom{10}{i}2^{i} = (1+2)^{10} = 3^{10}.$

Evaluate

$3^{10} = 59049.$

Therefore

$\frac{A}{10^{4}} = \frac{59049}{10000} = 5.9049.$

Locate the interval

$5.9049$ satisfies $4 \le 5.9049 < 7$, so it lies in $[4, 7)$ — option 3.

Exam trick: Whenever you see $\sum \binom{n}{i}c^{i}$, collapse it instantly to $(1+c)^{n}$ instead of summing terms. A quick sanity bound also nails it: $3^{10}>3^{9}=19683$ and $3^{10}<10^{5}$, so $A/10^4$ is between roughly $2$ and $10$, and $3^{10}\approx 59049$ pins it firmly inside $[4,7)$.

Note on the answer key: the preliminary key marked option 4. By the binomial theorem $A=\sum_{i=0}^{10}2^{i}\binom{10}{i}=(1+2)^{10}=3^{10}=59049$, so $\dfrac{A}{10^{4}}=5.9049\in[4,7)$. Correct answer: option 3. The preliminary key is an erratum.