SOLUTION: Hello I seem to be stuck at this problem. I'm not to sure how to write it as the n and n-1 are subscripts. Show that an = Sn � Sn-1

Question 346582: Hello I seem to be stuck at this problem.
I'm not to sure how to write it as the n and n-1 are subscripts.
Show that an = Sn � Sn-1
Found 2 solutions by Theo, Fombitz:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
You can use brackets, or you can use parentheses.
parentheses are easier to write.
brackets are more formal.
i'll use parentheses.

formula is a(n) = S(n) - S(n-1) where () means sub.

In general, S(n) = a(1) + a(2) + ........ a(n-1) + a(n)

In general S(n-1) = a(1) + a(2) + ........a (n-1)

If you subtract one from the other, you are left with a(n) because all of the a's from a(0) to a(n-1) cancel out.

a(1) - a(1) = 0
a(2) - a(2) = 0
a(n-1) - a(n-1) = 0
a(n) - 0 = a(n)

nothing left but a(n).

there's a more formal proof that can be done, but it's more complicated to go through and not worth the effort unless you absolutely have to do it that way.

for your purposes, this shows that it's true and should be sufficient.

you can then cover with examples:

let a1 = 5 and let d = 7 (d = common difference in arithmetic sequence).

s(7) = 5 + 12 + 19 + 26 + 33 + 40 + 47 = 182

s(8) = 5 + 12 + 19 + 26 + 33 + 40 + 47 + 54 = 236

s(8) - s(7) = 236 - 182 = 54

Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
S[n]=a[1]+a[2]+a[3]+...+a[n-2]+a[n-1]+[an]
.
.
S[n-1]=a[1]+a[2]+a[3]+...+a[n-2]+a[n-1]
.
.
S[n]-S[n-1]=(a[1]-a[1])+(a[2]-a[2])+(a[3]-a[3])+...+(a[n-2]-a[n-2])+(a[n-1]-a[n-1])+a[n]
S[n]-S[n-1]=(0)+(0)+(0)+...+(0)+(0)+a[n]
S[n]-S[n-1]=a[n]
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