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Ask for a mathematical sequence problem in Senior Two The first n terms of sequence an and Sn = 32n-n ^ 2 1. The general formula of an 2. Find the first n terms and TN of an (the first two are absolute values.)
① Using sn-sn-1 = an
An = 32n-n & sup2; - 32 (n-1) + (n-1) & sup2;
=33-2n
When n = 1
an=31=32-1
establish
②an≥0 n＜17
So the sum of the first 16 items is
（31+33-2n）*n/2
=(32-n)*n
After 16
An = 2n-331
The sum is (1 + 2n-33) * (n-16) / 2
=(n-16)²
Add the first 16 items and 16 & sup2; = 256
in summary
TN = 32-n * n ≤ 16 n is a positive integer
(n-16) & sup2; + 256 n > 16 n is a positive integer
Mathematics Series in senior two is very urgent
Let the sum of the first n terms of the arithmetic sequence {an} be Sn, A3 = 12, S12 > 0, S13 < 0
1) Find out the range of tolerance D
2) Point out which value of S1, S2, S3,. SN is the biggest and explain the reason
Detailed explanation, O (∩)_ Thank you
Because A3 = 12 = 3A1 + 3D
So a1 + D = 4
Because S12 = 12a1 + 66d is greater than 0
S13 = 13a1 + 78d < 0
So D is greater than 4 / 5 and less than 8 / 9
Given that both {an} and {BN} are arithmetic sequences, and A1 = 1, B1 = 4, a25 + B25 = 149, then the sum of the first 25 terms of {an + BN} is equal to?
Summation formula of arithmetic sequence
The sum of the first 25 terms of {an + BN} = (a1 + B1 + a25 + B25) * 25 / 2
=（a1+a25）*25/2+（b1+b25）*25/2
=（1+4+149）*25/2
Press the calculator for the rest~
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