prove SN- 2SN-1 + Sn-2 = d, for an AP where first term is a & common diff. is d
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S(n) = n/2 × ( 2a + (n-1)d) = (2an + d(n^2) - nd)/2
S(n-1) = (n-1)/2 × ( 2a + (n-2)d) =
( 2an + (n^2)d - 2a + 2d - 3nd)/2
S(n-2) = (n-2) × ( 2a + (n-3)d) =
(2an +(n^2)d - 5nd - 4a + 6d)/2
now filling in equation
S(n) - 2S(n-1) + S(n-2)
the calculation further is showed in pic
S(n-1) = (n-1)/2 × ( 2a + (n-2)d) =
( 2an + (n^2)d - 2a + 2d - 3nd)/2
S(n-2) = (n-2) × ( 2a + (n-3)d) =
(2an +(n^2)d - 5nd - 4a + 6d)/2
now filling in equation
S(n) - 2S(n-1) + S(n-2)
the calculation further is showed in pic
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Answer: check the attachment
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