There are seven numbers, the average of which is 18; after one number is removed, the average of the remaining six numbers is 19; after another number is removed Next to the question: the average of the remaining five numbers is 20 There should be formulas

The average of the first five numbers is 20, so the sum of the first five numbers is 100
The average of the first six numbers is 19, so the sum of the first six numbers is 114
So the sixth number is 114-100 = 14
The average of the first seven numbers is 18, so the sum of the first seven numbers is 126
So the seventh number is 126-114 = 12
So the product of the two numbers removed is 14 * 12 = 168

There are seven numbers, and their average is 18. After one number is removed, the average of the remaining six numbers is 19. After another number is removed, the average of the remaining five numbers is 20. What is the product of the two numbers removed?

There are seven numbers, the average of which is 18; after one number is removed, the average of the remaining six numbers is 19; after one number is removed, the remaining five numbers are 20
Find the product of two removed numbers

There are seven numbers, the average of which is 18. After one number is removed, the average of the remaining six numbers is 19. What is the number removed?

7×18-6×19
=126-114
=12

Given that the general term of the sequence {an} is an = (n - √ 98) / (n - √ 99), n ∈ natural number, then in the first 30 terms of the sequence {an}, the maximum term and the minimum term are ()
a10 a9
But after I took the derivative of this function, I found that it was simple minus, which should be A1 A30
Kneel down to ask the prawn to solve the problem

The answer is A10 A9
N ∈ natural number, not n - √ 98, N - √ 99 greater than 0

If the sum of the first n terms of the equal ratio sequence {an} is Sn, and S5 = 2, S10 = 6, then a16 + A17 + A18 + A19 + A20=______ ．

∵ S5 = 2, S10 = 6, ∵ A6 + A7 + A8 + A9 + A10 = 6-2 = 4, ∵ a1 + A2 + a3 + A4 + A5 = 2, ∵ Q5 = 2, ∵ a16 + A17 + A18 + A19 + A20 = (a1 + A2 + a3 + A4 + A5) Q15 = 2 × 23 = 16

Help solve several series of problems
1. Sequence 15 / 2, 24 / 5, 35 / 10, 48 / 17, 63 / 26 A general formula of is_____ .
2. Find a general formula of the following sequence
①3,5,9,17,33,……
②1,0,-1/3,0,1/5,0,-1/7,0,……
③4/5,1/2,4/11,2/7,……

1、（（n+3）^2)/(n^2+1)
2-1:2^n+1

Given that f (x) = 3-4 ^ x + 2xln2, the sequence {a subscript n} satisfies: - 1 / 2

At first glance, this problem should be the final type of senior high school mathematics. Hehe, it's exciting at first sight~
I use the tone of reminder to explain this problem, I think you should be smart enough ~!
The breakthrough of this problem is function. You can see that the equation of this function is so complex and strange. If you don't see it clearly, then this problem can't be solved
Methods: first of all, the concept of sequence is put aside, and the function f (x) = 3-4 ^ x + 2xln2 is proved in X

Known arrays: (11), (12, 21), (13, 22, 31), (14, 23, 32, 41) ，(1n，2n-1，3n-2，… ，n-12，n1)，… Note that the array is: (A1), (A2, A3), (A4, A5, A6) Then A2009 = 0___ ．

Let A2009 be the nth number in group M, then: 1 + 2 + 3 + +(M-1)＜20091+2+3+… +M ≥ 2009: M = 63, and 1 + 2 + 3 + +62 = 1953 ∵ 2009-1953 = 56 ∵ A2009 is the 56th number in group 63 ∵ A2009 is the 56th number in group 63 = 5663-56 + 1 = 7, so the answer is: 7

Each item of the arithmetic sequence {an} is positive, A1 = 3, the sum of the first n terms is Sn, {BN} is the arithmetic sequence, B1 = 1, and b2s2 = 64, b3s3 = 960
(1) Finding an and BN
(2) Sum: 1 / S1 + 1 / S2 + +1/Sn

First question:
Let the tolerance be D and the common ratio be Q
Then B2 = q, S2 = a1 + A2 = 3 + 3 + D = 6 + D, B3 = q ^ 2, S3 = a1 + A2 + a3 = 3 + 3 + D + 3 + 2D = 9 + 3D
B2s2 = q (6 + D) = 64, b3s3 = q ^ 2 × (9 + 3D) = 960, namely Q ^ 2 × (3 + D) = 320
From Q (6 + D) = 64, we get 6 + D = 64 / Q, then 3 + D = (64 / Q) - 3, so Q ^ 2 × (3 + D) = q ^ 2 × (64 / Q) - 3Q ^ 2 = 64q-3q ^ 2 = 320, so q = 40 / 3 or q = 8, but when q = 40 / 3, D is negative, so it is discarded
So q = 8, d = 64 / 8-6 = 2
So an = 3 + 2 (n-1) = 2n + 1, BN = 1 × 8 ^ (n-1) = 8 ^ (n-1)
Second question:
Sn=(3+2n+1)×n/2=n×(n+2)
So 1 / Sn = 1 / [n × (n + 2)] = (1 / 2) × (1 / n-1 / (n + 2))
So 1 / S1 + +1/Sn=1/2×(1/1-1/3+1/3-1/5+…… +1/n-1/(n+2))=(1/2)×(1-1/(n+2))=(n+1)/(2n+4)

There are seven numbers, the average of which is 18; after one number is removed, the average of the remaining six numbers is 19; after another number is removed Next to the question: the average of the remaining five numbers is 20 There should be formulas