Giving Presents – Geometric Sequences

Answer: $4^{n-1}$ .

Since it follows that on the first day, only $1$ person receives presents and gives it to $3$ other people on the second day, and so forth, we get the sequence (of the total number of recipients of presents): $1, 4, 16, 64, …$

Further, we can denote each term in the pattern as $4\times 4\times 4\times ...\times 4$ , with the expression consisting of $n$ terms, with $n$ being the day number. Algebraically, given $x$ people, the $x$ people will give presents to $3x$ people. Since we are finding the total number of recipients of presents, the next term in the sequence must be $x+3x=4x$ . This means each term is four times the previous, therefore we get $4^{n}$ . Note that the index must be $n-1$ instead of $n$ because the first term is 1 ( $4^{1-1}$ ), the second term is 4 ( $4^{2-1}$ ), and so forth, and therefore the pattern determined is $4^{n-1}$ .

Answer: $\displaystyle\{^{1,\ if\ n\,=\,1}_{3\,\times\,4^{n\,-\,2},\ if\ n\,>\,1}$ .

Given that we are to find the new recipients of presents from the day, we don’t need to count the people who have already obtained presents. Thus, we get the sequence: 1 for the first day, as one person obtains a present; 3 for the second day, as the person from day one gives a present to three other people respectively; 12 for the third day, as the people from days one and two (a total of 4) give presents to three other people ( $4\times 3=12$ ), and so on: 1, 3, 12, 48, 192, …

Note that adding all previous terms and the current term in the sequence gives the current term in the sequence from part 1.From this sequence, we see that the second term is triple the first; the third quadruple the second; the fourth quadruple the third; and so on, quadrupling each term following onwards. This also involves exponents, and we can derive the entire sequence from piecewise function/pattern:

$\displaystyle\{^{1,\ if\ n\,=\,1}_{3\,\times\,4^{n\,-\,2},\ if\ n\,>\,1}$

Answer: $18$

Given the pattern $\displaystyle\{^{1,\ if\ n\,=\,1}_{3\,\times\,4^{n\,-\,2},\ if\ n\,>\,1}$ from part 2, we can simply evaluate the bottom part ( $3\times4^{n\,-\,2}$ ) in an inequality: $3\times4^{n\,-\,2}\ge 10,000,000,000$ , and solve for $n$ .

$4^{n\,-\,2}\ge{10,000,000,000\over{3}}$

$\log_{4}{10,000,000,000\over 3}\leq n-2$

$15.82\leq n-2$ (approximated to nearest hundredth)

$n\ge 17.82$ , and since we are finding the first day $n$ that the number of new recipients who receive presents on the day exceeds 10 billion, we just need to find the smallest natural number $n$ satisfying the inequality, which is $\boxed{18}$ .