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Difference between revisions of "1967 AHSME Problems/Problem 39"

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== Problem ==
 
== Problem ==
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Given the sets of consecutive integers <math>\{1\}</math>,<math> \{2, 3\}</math>,<math> \{4,5,6\}</math>,<math> \{7,8,9,10\}</math>,<math>\; \cdots \; </math>, where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set.  Let <math>S_n</math> be the sum of the elements in the nth set.  Then <math>S_n</math> equals:
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Given the sets of consecutive integers <math>\{1\}</math>,<math> \{2, 3\}</math>,<math> \{4,5,6\}</math>,<math> \{7,8,9,10\}</math>,<math>\; \cdots \; </math>, where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set.  Let <math>S_n</math> be the sum of the elements in the nth set.  Then <math>S_{21}</math> equals:
  
 
<math>\textbf{(A)}\ 1113\qquad
 
<math>\textbf{(A)}\ 1113\qquad
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== Solution ==
 
== Solution ==
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The last element of the 21st set is <math>\frac{21\times 22}{2}=231</math>, and hence the first element of the 21st set is <math>211</math>. So <math>S_{21}=\frac{231\times 232}{2}-\frac{210\times 211}{2}</math>, hence our answer is <math>\fbox{B}</math>
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The last element of the 21st set is <math>\frac{21\times 22}{2}=231</math>, and hence the first element of the 21st set is <math>211</math>.
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So <math>S_{21}=\frac{231\times 232}{2}-\frac{210\times 211}{2}=4641</math>, hence our answer is <math>\fbox{B}</math>
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1967|num-b=38|num-a=40}}   
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{{AHSME 40p box|year=1967|num-b=38|num-a=40}}   
  
 
[[Category: Intermediate Algebra Problems]]
 
[[Category: Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:40, 16 August 2023

Problem

Given the sets of consecutive integers $\{1\}$,$\{2, 3\}$,$\{4,5,6\}$,$\{7,8,9,10\}$,$\; \cdots \;$, where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set. Let $S_n$ be the sum of the elements in the nth set. Then $S_{21}$ equals:

$\textbf{(A)}\ 1113\qquad \textbf{(B)}\ 4641 \qquad \textbf{(C)}\ 5082\qquad \textbf{(D)}\ 53361\qquad \textbf{(E)}\ \text{none of these}$

Solution

The last element of the 21st set is $\frac{21\times 22}{2}=231$, and hence the first element of the 21st set is $211$. So $S_{21}=\frac{231\times 232}{2}-\frac{210\times 211}{2}=4641$, hence our answer is $\fbox{B}$

See also

1967 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 38 		Followed by
Problem 40
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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