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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 12:34, 30 April 2017
Problem 25
The sum of all numbers of the form , where takes on integral values from to is:
Solution
The sum of the odd integers from to is . However, in this problem, the sum is instead , starting at rather than . To rewrite this, we note that is less than for every we add, so for 's, we subtract , giving us ,which factors as .
See Also
1956 AHSME (Problems • Answer Key • Resources) | ||
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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