Binomial nested sum
r = 0 ∑ n ( − 1 ) r ( r n ) − 1
If n is an an odd positive integer, find the value of this sum.
The answer is 0.
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4 solutions
is it valid to say that the summation is zero since by the definition of combination: ( r n ) = r ! ( n − r ) ! n ! if 0 ≤ r ≤ n , or 0 if r > n or r < 0 , for any r, n ∈ Z with n ≥ 0 .
We know that ( r n ) is always an integer. Yet its inverse cannot be in the form n 1 , n ∈ N . Hence, ( r n ) − 1 is zero.
Is there any error with this reasoning? By the way, i got it right.. :D
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jeffer dave cagubcob --it is hard to say that u that --
ur above explanation is wrong.
this might or is true when n-->infinite .
not when "n" is small .
hence the above problem can be solved by converting the even terms into odd once
Thus they cancel out each other.
What if n is even?
If we set n equal to 1 , then it is clear that the answer is 0 .
Just split the series in half, and recognize that (because of pascal's triangle's symmetry) the right hand side is the same as the left hand side except for the sense (because of the powers of -1 alternating between 1 and 1). Alternatively (I did this but almost immediately recognized the formal proof, so I knew the proper solution/proof before I answered) you could just try it for 1 (or any positive integer, but 1 is the easiest) as this question was poorly constructed. You can also get a sense that the answer would be 0, more from gut feelings and conjecture, because summations could be 0 if it cancels out all the time and the odd integer part would suggest a canceling out as it would probably be related to have an even amount of elements (so one elements always has it's additive group inverse) to cancel out to 0 (just in case you were looking for another way to start of a question like this beside looking at cases like I did).
Anything to the power of -1 is always 0
Are you sure about that?
That is not correct, as 0.5 to the power of -1 is 2 (try it on a calculator)
Recall that ( r n ) = ( n − r n ) . Since n is odd, n − r has the opposite parity as r , so
( − 1 ) r ( r n ) − 1 = − ( − 1 ) n − r ( n − r n ) − 1 .
Thus, when we pair the terms in this way, each pair sums to 0, so the summation is 0.