Solving the alternating sum of binomial coefficients using telescoping series

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The discussion focuses on proving that the alternating sum of binomial coefficients equals zero for any integer n. The initial approach involves using proof by induction, starting with a base case of n=1. Participants suggest that the problem can be approached as a telescoping series, emphasizing the importance of the binomial coefficients in the formulation. The user seeks clarity on the layout of their expression and assistance in refining their proof. Overall, the conversation revolves around understanding and solving the alternating sum through mathematical techniques.
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I havn't done this in a long time! And apparently I should know this easy, it sort of looks like a proof by induction, which I havn't done before and I am frantically trying to learn!

Show that for each integer n the alternating sum of binomial coefficients:

1 - (n) + ... + (-1)^k(n) + ... + (-1)^(n-1)( n ) + (-1)^n
...(1)......(k)...... ...(n-1)
is zero. What is the value of the sum

(so what I've done here is started with an "inductive basis of n=1 which kind of suggests it goes to zero but without the appropriate conciseness)

1 + (n) + ... + (n) + ... + ( n ) +1
...(1)...(k)...(n-1)

I understand the layout is a bit rubbish but I hope you can fathom it!
Any help would be greatly appreciated!

UPDATE! After a bit of research, am I correct in assuming this is a telescoping series?
 
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"Binomial" is the key word here. Try to find a clever way to manufacture what you see in front of you by expanding some expression.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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