Difference between revisions of "1962 AHSME Problems/Problem 12"
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By the Binomial Theorem, the sum of the last three coefficients is | By the Binomial Theorem, the sum of the last three coefficients is | ||
<math>\binom{6}{2}-\binom{6}{1}+\binom{6}{0}=15-6+1=\boxed{10 \textbf{ (C)}}</math>. | <math>\binom{6}{2}-\binom{6}{1}+\binom{6}{0}=15-6+1=\boxed{10 \textbf{ (C)}}</math>. | ||
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| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1962|before=Problem 11|num-a=13}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 22:15, 3 October 2014
Problem
When 
is expanded the sum of the last three coefficients is:
Solution
This is equivalent to 
Its expansion has 7 terms, whose coefficients are the same as those of 
.
By the Binomial Theorem, the sum of the last three coefficients is
.
See Also
| 1962 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
