In the sequence {an}, A1 = 1, an-1 = an + 2N-1 (n belongs to positive integer), then an The general formula of is?

In the sequence {an}, A1 = 1, an-1 = an + 2N-1 (n belongs to positive integer), then an The general formula of is?

∵an-a(n-1)=1-2n
∴a(n-1)-a(n-2)=1-2(n-1)
a(n-2)-a(n-3)=1-2(n-2)
……
a3-a2=1-2×3
a2-a1=1-2×2
The sum of the above results in
an-a1=[1-2n]+[1-2(n-1)]+[1-2(n-2)]+…… +[1-2×3]+[1-2×2]
=(n-1)-2[n+(n-1)+(n-2)…… +3+2]
The numbers in braces are arithmetic sequences
∴an-a1=(n-1)-2[(n²+n-2)/2]=n-1-(n²+n-2)=n-1-n²-n+2=1-n²
∵a1=1
∴an=2-n²

In known sequence {an}, A1 = 1 / 2, and a (n + 1) = an / 2 + (2n + 3) / 2 ^ (n + 1), n is a positive integer, find an

This problem is still using the recursive method, using its own law to find an, the teacher has demonstrated in class, it is not difficult to think about it seriously
an=(n²+2n-2)/2^n

If the sequence {an} is a positive integer for any m, and satisfies am + n = am + an, and A3 = 8, then A10 =?

Am+n=Am+An
Let n = 1 have
Am+1=Am+A1
The sequence am is an arithmetic sequence, the first term is A1, and the tolerance is A1, so that
Am=A1+(m-1)A1=nA1
A1=A3/3=8/3
A10=10A1=80/3