Given the A1 = 2 a n + 1 - a n = 3 of the sequence {an}, the general term formula of the sequence is

Given the A1 = 2 a n + 1 - a n = 3 of the sequence {an}, the general term formula of the sequence is

From the meaning of the title, a (n + 1) - A (n) = 3, a (1) = 2,
Then this is the arithmetic sequence with the first term of 2 and the tolerance of D = 3;
The general formula is: a (n) = a (1) + (n-1) d = 2 + (n-1) 3 = 3n-1
That is, a (n) = 3n-1

Given that the sum of the first n terms of sequence {a (n)} is s (n), A1 = 1, a (n + 1) = 1 / 3Sn, the general term formula of sequence {a (n)} is obtained

a(n+1)=1/3Sn
a(n)=1/3S(n-1)
a(n+1)-an=1/3(Sn-S(n-1))=1/3an
therefore
a(n+1)=4/3an
So an is an equal ratio sequence with the first term of 1 and the common ratio of 4 / 3
an=（4/3）^(n-1)

1 ^ 4 + 2 ^ 4 + 3 ^ 4 +What's the formula for ^ n?

1^4+2^4+3^4+4^4+…… +n^4
=n(n+1)(2n+1)(3n^2+3n-1)/30
prove:
(n+1)^5-n^5=5n^4+10n^3+10n^2+5n+1
n^5-(n-1)^5=5(n-1)^4+10(n-1)^3+10(n-1)^2+5(n-1)+1
……
2^5-1^5=5*1^4+10*1^3+10*1^2+5*1+1
Add it all up
(n+1)^5-1^5=5*(1^4+2^4+3^4+4^4+…… +n^4)+10*(1^3+2^3+3^3+4^3+…… +n^3)+10*(1^2+2^2+3^2+4^4+…… +n^2)+5*(1+2+3+4+…… +n)+n
Because 1 ^ 3 + 2 ^ 3 + 3 ^ 3 + 4 ^ 3 + +n^3=[n(n+1)/2]^2
1^2+2^2+3^2+4^4+…… +n^2=n(n+1)(2n+1)/6
1+2+3+4+…… +n=n(n+1)/2
So 1 ^ 4 + 2 ^ 4 + 3 ^ 4 + 4 ^ 4 + +n^4
={[(n+1)^5-1^5]-10*[n(n+1)/2]^2-10*n(n+1)(2n+1)/6-5*n(n+1)/2-n}/5
=n(n+1)(2n+1)(3n^2+3n-1)/30
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