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Does anyone know how to derive the formula Tn = n/2 (a+1)?
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Does anyone know how to derive the formula Tn = n/2 (a+1)?
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Answer #1
| 10/02 2014 06:39
More information is needed.
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
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Answer #1| 10/02 2014 06:39