1967 AMC 12 Problem 39

Question

Given the sets of consecutive integers

\{1\},

\{2,3\},

\{4,5,6\},

\{7,8,9,10\},

\ldots,

where each set contains one more element than the preceding one, and where the first element of each succeeding set is one more than the last element of the preceding set. Let

S_n

be the sum of the elements in the

n

th set. Then

S_{21}

equals:

1113

4641

5082

53361

none of these

Answer: B

Concepts:arithmetic sequence triangular number summation

Difficulty rating: 1500

Small hint:

The last number in the

n

th set is

\frac{n(n+1)}{2}

Big hint:

Sum the

n

consecutive integers ending at that number

Solution:

The

n

th set ends at

\frac{n(n+1)}{2}

and contains

n

consecutive integers. Its sum is

\begin{aligned} S_n &=n\left(\frac{n(n+1)}2\right)\\ &\quad-\frac{n(n-1)}2\\ &=\frac{n(n^2+1)}2. \end{aligned}

Thus

S_{21}=\frac{21(442)}{2}=4641.

Therefore, the correct answer is B.