Explain why Sn=na1 n(n−1)2d s sub n , equals n , eh sub 1 , plus . fraction n . open , n minus 1 , close , over 2 end fraction . d makes sense by explaining how you can extract each of na1 n eh sub 1 and n(n−1)2d fraction n . open , n minus 1 , close , over 2 end fraction . d from the sum a1 (a1 d) (a1 2d) ⋯ (a1 (n−1)d).
