Difference between revisions of "1956 AHSME Problems/Problem 25"

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==Solution==
 
==Solution==
 
The sum of the odd integers <math>2k-1</math> from <math>1</math> to <math>n</math> is <math>n^2</math>. However, in this problem, the sum is instead <math>2k+1</math>, starting at <math>3</math> rather than <math>1</math>. To rewrite this, we note that <math>2k-1</math> is <math>2</math> less than <math>2k+1</math> for every <math>k</math> we add, so for <math>n</math> <math>k</math>'s, we subtract <math>2n</math>, giving us <math>n^2+2n</math>,which factors as <math>n(n+2) \implies \boxed{\text{(C)} n(n+2)}</math>.
 
The sum of the odd integers <math>2k-1</math> from <math>1</math> to <math>n</math> is <math>n^2</math>. However, in this problem, the sum is instead <math>2k+1</math>, starting at <math>3</math> rather than <math>1</math>. To rewrite this, we note that <math>2k-1</math> is <math>2</math> less than <math>2k+1</math> for every <math>k</math> we add, so for <math>n</math> <math>k</math>'s, we subtract <math>2n</math>, giving us <math>n^2+2n</math>,which factors as <math>n(n+2) \implies \boxed{\text{(C)} n(n+2)}</math>.
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==See Also==
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{{AHSME box|year=1956|num-b=24|num-a=26}}
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{{MAA Notice}}

Latest revision as of 12:34, 30 April 2017

Problem 25

The sum of all numbers of the form $2k + 1$, where $k$ takes on integral values from $1$ to $n$ is:

$\textbf{(A)}\ n^2\qquad\textbf{(B)}\ n(n+1)\qquad\textbf{(C)}\ n(n+2)\qquad\textbf{(D)}\ (n+1)^2\qquad\textbf{(E)}\ (n+1)(n+2)$

Solution

The sum of the odd integers $2k-1$ from $1$ to $n$ is $n^2$. However, in this problem, the sum is instead $2k+1$, starting at $3$ rather than $1$. To rewrite this, we note that $2k-1$ is $2$ less than $2k+1$ for every $k$ we add, so for $n$ $k$'s, we subtract $2n$, giving us $n^2+2n$,which factors as $n(n+2) \implies \boxed{\text{(C)} n(n+2)}$.

See Also

1956 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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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