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anonymous
 4 years ago
Use the Triangle Inequality and Mathematical Induction to show that\[\left\sum_{k=1}^{n}a_k\right\leq\sum_{k=1}^{n}a_k.\]Well, from the Triangle Inequality, we know that\[\left\sum_{k=1}^{n}a_k\right=a_1+a_2+...+a_n\leqa_1+a_2+...+a_n=\sum_{k=1}^{n}a_k,\]and by Mathematical Induction, we see that\[\left\sum_{i=1}^{1}a_i\right=a_1=\sum_{i=1}^{1}a_i,\]assume it's true for \(k\), and see that\[\left\sum_{i=1}^{k+1}a_i\right=a_1+a_2+...+a_{k+1}\leqa_1+a_2+...+a_{k+1}=\sum_{i=1}^{k+1}a_i,\]as required.
anonymous
 4 years ago
Use the Triangle Inequality and Mathematical Induction to show that\[\left\sum_{k=1}^{n}a_k\right\leq\sum_{k=1}^{n}a_k.\]Well, from the Triangle Inequality, we know that\[\left\sum_{k=1}^{n}a_k\right=a_1+a_2+...+a_n\leqa_1+a_2+...+a_n=\sum_{k=1}^{n}a_k,\]and by Mathematical Induction, we see that\[\left\sum_{i=1}^{1}a_i\right=a_1=\sum_{i=1}^{1}a_i,\]assume it's true for \(k\), and see that\[\left\sum_{i=1}^{k+1}a_i\right=a_1+a_2+...+a_{k+1}\leqa_1+a_2+...+a_{k+1}=\sum_{i=1}^{k+1}a_i,\]as required.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Does this seem sound?

LifeIsADangerousGame
 4 years ago
Best ResponseYou've already chosen the best response.0I agree with Hershey_Kisses...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0holy cow! that looks like.....holy cow that unbearable!
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