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Arithmetic progression

Derive s=n/2(a+l)
Original post by Bushgeorge
Derive s=n/2(a+l)


Can you first prove that 1+2+3+...+n=12n(n+1)\displaystyle 1+2+3+...+n = \frac{1}{2}n(n+1) ?
you can split the progression into pairs, each pair having the same sum. then you just multiply the sum of a pair by the number of pairs.
Reply 3
Original post by RDKGames
Can you first prove that 1+2+3+...+n=12n(n+1)\displaystyle 1+2+3+...+n = \frac{1}{2}n(n+1) ?


I don't know that is why I need an answer

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