2+3+4+....+n 怎麼用等差級數來算?

2018-08-17 5:49 am

回答 (2)

2018-08-17 4:02 pm
sn-1= (1+2+ 3+ 4+…+ n )-1
a1=1 d=1 n=1+(n-1)1
sn=[(a1+n)n]/2
sn=[(1+n)n]/2
sn=(n^2+n)/2
sn-1=(n^2+n)/2-1
2018-08-17 1:09 pm
2+3+4+....+n 怎麼用等差級數來算?
Sol
A= 2+ 3+ 4+…+ n
A= n+(n-1)+(n-2)+…+ 2
-----------------------------------------------
2A=(n+2)+(n+2)+ (n+2)+…+ (n+2)
=(n-2+1)(n+2)
=(n-1)(n+2)
A=(n-1)(n+2)/2=(n^2+n-2)/2
or
2+3+4+…+n
=(1+2+3+…+n)-1
=n(n+1)/2-1
=(n^2+n)/2-1
=(n^2+n-2)/2


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