Is the expression given by:
n
2 - (n-1)
2+(n-2)
2+.....+ (-1)
n-2*2
2+(-1)
n-1*1
2
a triangular number for every value of n?
Prove your assertion.
Suppose n is even.
Split up the series into pairs. There will be n/2 new terms:
2n-1 + 2n-5 + 2n-9 + ... + 2n - (2n-3)
This evaluates to X - Y, where
X is (n/2)(2n) = n^2
Y is 1 + 5 + 9 + ... + (2n-3) which is a geometric series which sums to:
(n/2)*((2n-2)/2) = n(n-1)/2
Our sum is X - Y = n^2 - (n(n-1)/2)
which evaluates to
n(n+1)/2If instead n is odd, then consider m = n-1, or n = m+1.
m is even, so we know that all but the first term sum to m(m+1)/2 except with the sign flipped.
The first term is n^2 or (m^2 + 2m + 1)
So, for odd n, the sum in terms if m is:
m^2 + 2m + 1 - (m(m+1)/2) = (m+1)(m+2)/2 =
n(n+1)/2
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Posted by Larry
on 2023-06-24 11:30:34 |
solution