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1996 AHSME Problems/Problem 24

Revision as of 21:33, 6 October 2019 by Zhuqingzhang (talk | contribs) (→‎Solution)

Problem

The sequence $1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,\ldots$ consists of $1$’s separated by blocks of $2$’s with $n$ $2$’s in the $n^{th}$ block. The sum of the first $1234$ terms of this sequence is

$\text{(A)}\ 1996\qquad\text{(B)}\ 2419\qquad\text{(C)}\ 2429\qquad\text{(D)}\ 2439\qquad\text{(E)}\ 2449$

Solution

The sum of the first $1$ numbers is $1$

The sum of the next $2$ numbers is $2 + 1$

The sum of the next $3$ numbers is $2 + 2 + 1$

In general, we can write "the sum of the next $n$ numbers is $1 + 2(n-1)$", where the word "next" follows the pattern established above.

Thus, we first want to find what triangular numbers $1234$ is between. By plugging in various values of $n$ into $f(n) = \frac{n(n+1)}{2}$, we find:

$f(50) = 1275$

$f(49) = 1225$

Thus, we want to add up all those sums from "next $1$ number" to the "next $49$ numbers", which will give us all the numbers up to and including the $1225^{th}$ number. Then, we can manually tack on the remaining $2$s to hit $1234$.

We want to find:

$\sum_{n=1}^{49} 1 + 2(n-1)$

$\sum_{n=1}^{49} 2n - 1$

$\sum_{n=1}^{49} 2n - \sum_{n=1}^{49} 1$

$2 \sum_{n=1}^{49} n - 49$

$2\cdot \frac{49\cdot 50}{2} - 49$

$49^2$

$2401$

Thus, the sum of the first $1225$ terms is $2401$. We have to add $9$ more $2$s to get to the $1234^{th}$ term, which gives us $2419$, or option $\boxed{B}$.

Note: If you notice that the above sums form $1 + 3 + 5 + 7... + (2n-1) = n^2$, the fact that $49^2$ appears at the end should come as no surprise.

Solution 2

We note that the number of elements in each block are

$2, 3, 4,$ and so on.

Let $n$ be the number of blocks up to $1234$ terms.

We have $\frac {2+n+1}{2} \cdot (n)=1234$

Solving for $n$, we get about $48.2$.

We have $48$ full blocks and $1$ partial block. We find that there are a total of $49$ $1$'s

Now, we change every number in the sequence to $2$, and then add. We get $2468$. Since we added $49$ by changing all $49$ $1$'s to $2$, we must subract that $49$. Giving us

$2468-49=2419$ $\boxed{B}$

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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