# Does anyone know how to derive the formula Tn = n/2 (a+1)?

• Does anyone know how to derive the formula Tn = n/2 (a+1)?

Answer #1 | 10/02 2014 06:39
Positive: 96 %
Answer #2 | 10/02 2014 09:11
Generally, Tn = 1 + 2 + 3+ ....... + (n-1) + n --------> eqn 1 Tn = n + (n-1) +(n-2)+......+ 2 + 1 ---------> eqn 2 adding eqn 1 and eqn 2 Tn + Tn = (1+n) + [2+(n-1)] + [3+ (n-2)] +.....+[(n-1)+2] + (n+1) 2Tn = (1+n) + (n + 1 )+ (n + 1 )+ .....+ (n + 1 )+ (n + 1 ) All the n pair-sums are equal to (n+1) 2 Tn = n(n+1) Tn = n(n+1)/2
Positive: 90 %

[Python] numpy/scipy: correlation; Robert. ... (n-2)/(1-r**2) ) ... I guess the formula for r does not change (much) ...
Positive: 96 %
... [n] = (2 n /n+1) * sin[(*n)/2 u[n] ... Does anyone have any recommendations on how to do this? ... n ?? 1 tn??1estu(t)
Positive: 91 %
... Reference desk/Archives/Mathematics/2013 ... I thought that formula I just found was the way to derive ... It does imply that the periods T1...TN ...
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