Difference between revisions of "1957 AHSME Problems/Problem 30"

(created solution page)
Line 1: Line 1:
<math>c=k=1</math>
+
== Problem ==
Option D
+
 +
The sum of the squares of the first n positive integers is given by the expression <math>\frac{n(n + c)(2n + k)}{6}</math>,
 +
if <math>c</math> and <math>k</math> are, respectively:
 +
 
 +
<math>\textbf{(A)}\ {1}\text{ and }{2} \qquad \textbf{(B)}\ {3}\text{ and }{5}\qquad \textbf{(C)}\ {2}\text{ and }{2}\qquad\textbf{(D)}\ {1}\text{ and }{1}\qquad\textbf{(E)}\ {2}\text{ and }{1}    </math>
 +
 
 +
== Solution ==
 +
<math>\boxed{\textbf{(D) }1 \text{ and } 1}</math>.
 +
 
 +
== See Also ==
 +
{{AHSME 50p box|year=1957|num-b=29|num-a=31}}
 +
{{MAA Notice}}
 +
[[Category:AHSME]][[Category:AHSME Problems]]

Revision as of 17:59, 25 July 2024

Problem

The sum of the squares of the first n positive integers is given by the expression $\frac{n(n + c)(2n + k)}{6}$, if $c$ and $k$ are, respectively:

$\textbf{(A)}\ {1}\text{ and }{2} \qquad \textbf{(B)}\ {3}\text{ and }{5}\qquad \textbf{(C)}\ {2}\text{ and }{2}\qquad\textbf{(D)}\ {1}\text{ and }{1}\qquad\textbf{(E)}\ {2}\text{ and }{1}$

Solution

$\boxed{\textbf{(D) }1 \text{ and } 1}$.

See Also

1957 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 29
Followed by
Problem 31
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 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png