A mathematical sequence problem, a_ (n+1)=1/[3^(n+1)] Because a_ (n+1)=1/[3^(n+1)] So an = 1 / (3 ^ n)

A mathematical sequence problem, a_ (n+1)=1/[3^(n+1)] Because a_ (n+1)=1/[3^(n+1)] So an = 1 / (3 ^ n)

Let k = n + 1, then a_ (n + 1) = 1 / [3 ^ (n + 1)] Yes
a_ k =1/(3^k)
Then change K into n
a_ n=1/(3^n).

A series of mathematical problems, education!
Let {an} be an arithmetic sequence, {BN} be an arithmetic sequence with positive terms, and a1 + A6 + a11 = 4, B6 = A6, then lgb1 + lgb11=

a1+a6+a11=4
3a6=4
a6=4/3
b6=4/3
gb1+lgb11=lg(b1b11)=lgb6^2=2lgb6=2lg(4/3)=4lg2-2lg3

Sequence 1 / (1 ^ 2 + 2) + 1 / (2 ^ 2 + 4) + 1 / (3 ^ 2 + 6) + Sum of the top 18 items

This is the general formula of the denominator of each item: therefore, the sum of 18 items is the following formula: the original formula = 1 / 1 * (1 + 2) + 1 / 2 * (2 + 2) + 1 / 3 * (3 + 2) + 1 / 4 * (4 + 2) +. + 1 / 18 * (18 + 2) = 1 / 2 {(1-1 / 3) + (1 / 2)

Sequence (1 13:10:42)
Given that the sequence {an} satisfies A1 = 0, an + 1 + Sn = N2 + 2n (n belongs to n *), where SN is the sum of the first n terms of {an}, find the general term formula of the sequence

a(n+1)+Sn=S(n+1)=n^2+2n
S(n+1)+1=(n+1)^2
Sn=n^2-1 .(n>1)
S(n+1)-Sn=2n+1=a(n+1)
an=2n-1...(n>1)
an=0...(n=1)