if the sum of an arithmetic series S_n=n^2+2n find a formula for T_n, ie the nth term of the series
MC Squidge BOS' Apex Predator Joined May 13, 2008 Messages 267 Location none Gender Male HSC 2009 Mar 26, 2009 #1 if the sum of an arithmetic series S_n=n^2+2n find a formula for T_n, ie the nth term of the series
cookiesandcream ... Joined Oct 26, 2008 Messages 26 Gender Female HSC 2009 Mar 26, 2009 #2 Tn = 5/2 + (n-1) 5/2 i think?
MC Squidge BOS' Apex Predator Joined May 13, 2008 Messages 267 Location none Gender Male HSC 2009 Mar 26, 2009 #3 plz explain
D Drongoski Well-Known Member Joined Feb 22, 2009 Messages 4,255 Gender Male HSC N/A Mar 26, 2009 #4 MC Squidge said: if the sum of an arithmetic series S_n=n^2+2n find a formula for T_n, ie the nth term of the series Click to expand... Tn = Sn - Sn-1 = (n2 + 2n) - ( (n-1)2 + 2(n-1)) = 2n + 1 Is that right ? Last edited: Mar 27, 2009
MC Squidge said: if the sum of an arithmetic series S_n=n^2+2n find a formula for T_n, ie the nth term of the series Click to expand... Tn = Sn - Sn-1 = (n2 + 2n) - ( (n-1)2 + 2(n-1)) = 2n + 1 Is that right ?
D Drongoski Well-Known Member Joined Feb 22, 2009 Messages 4,255 Gender Male HSC N/A Mar 27, 2009 #5 Drongoski said: Tn = Sn - Sn-1 = (n2 + 2n) - ( (n-1)2 + 2(n-1)) = 2n + 1 Click to expand... Is this answer correct ??
Drongoski said: Tn = Sn - Sn-1 = (n2 + 2n) - ( (n-1)2 + 2(n-1)) = 2n + 1 Click to expand... Is this answer correct ??
MC Squidge BOS' Apex Predator Joined May 13, 2008 Messages 267 Location none Gender Male HSC 2009 Mar 27, 2009 #6 Drongoski said: Is this answer correct ?? Click to expand... dunno. dont have a solution
A alakazimmy Member Joined May 6, 2006 Messages 71 Gender Male HSC 2007 Mar 27, 2009 #7 Drongoski said: Tn = Sn - Sn-1 = (n2 + 2n) - ( (n-1)2 + 2(n-1)) = 2n + 1 Is that right ? Click to expand... Yup. T1 = 3. Tn = 2n + 1 So summing all those terms is equivalent to the first term, plus the last term, multiplied by: n/2 So, we have (2n + 4)*n/2 = n2 + 2n
Drongoski said: Tn = Sn - Sn-1 = (n2 + 2n) - ( (n-1)2 + 2(n-1)) = 2n + 1 Is that right ? Click to expand... Yup. T1 = 3. Tn = 2n + 1 So summing all those terms is equivalent to the first term, plus the last term, multiplied by: n/2 So, we have (2n + 4)*n/2 = n2 + 2n