Difference between revisions of "1970 Canadian MO Problems/Problem 9"
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since <math>(s+t)\equiv (s-t)\;(mod\;2)</math>, then <math>(-1)^{s+t}=(-1)^{s-t}</math> and our expression reduces to: | since <math>(s+t)\equiv (s-t)\;(mod\;2)</math>, then <math>(-1)^{s+t}=(-1)^{s-t}</math> and our expression reduces to: | ||
− | <math>f(s+t)-f(s-t)=\frac{2(s+t)^2-2(s-t)^2}{8} | + | <math>f(s+t)-f(s-t)=\frac{2(s+t)^2-2(s-t)^2}{8}=\frac{(s+t)^2-(s-t)^2}{4}</math> |
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<math>f(s+t)-f(s-t)=\frac{s^2+2st+t^2-s^2+2st-t^2}{4}=\frac{4st}{4}=st</math> | <math>f(s+t)-f(s-t)=\frac{s^2+2st+t^2-s^2+2st-t^2}{4}=\frac{4st}{4}=st</math> |
Latest revision as of 22:31, 27 November 2023
Problem 9
Let be the sum of the first terms of the sequence a) Give a formula for .
b) Prove that where and are positive integers and .
Solution
Part a):
Using in the formula we can have:
Therefore,
Part b):
since , then and our expression reduces to:
Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.