Note CN = an × BN, find the first n terms and TN of sequence {CN}, online and so on! an=4n-11 bn=2^n-1

Cn=(4n-11)(2^n-1)=(4n-11)2^n-4n+11
Let Sn = (4n-11) 2 ^ n = - 7 * 2-3 * 2 ^ 2 + 2 ^ 3 '+ (4n-11) 2 ^ n
Then 2Sn = (4n-11) 2 ^ (n + 1) = - 7 * 2 ^ 2``````````
2Sn-Sn=``````
That is to find CN
Finally, find TN

Let the general term formula of sequence {an} be 2 ^ n, the general term formula of sequence {BN} be 2N-1, and given sequence {CN} = BN / an, find the first n terms and TN of sequence CN

There are many ways to solve the problem, which can also be solved by using the integral of higher mathematics
Here's a simple way:
Tn=1/2+3/4+5/8+… +Formula (2n-1) / 2 ^ n a
2Tn=1+3/2+5/4+… +Formula (2n-1) / 2 ^ (n-1) B
B minus a
Tn=1+2/2+2/4+2/8+… +2/2^(n-1)-(2n-1)/2^n=2+1/2+1/4+… +1/2^(n-2)-(2n-1)/2^n
=3-(2n+3)/2^n
For example
T1=3-(2+3)/2=1/2=1/2
T2=3-(4+3)/4=5/4=1/2+3/4
T3=3-(6+3)/8=15/8=4/8+6/8+5/8=1/2+3/4+5/8

Given the sequence 1,2x, 3x ^ 2,..., NX ^ n-1 (x ≠ 0), find the sum of the first n terms

It's dislocation subtraction, not dislocation addition
S =1+2x+3x^2+4x^3+5x^4+…… +nx^(n-1) (1)
x=0,S=1;x=1,S=1+2+3+…… n=n(n+1)/2
xS=x+2x^2+3x^2…… +(n-1)x^n+nx^n (2)
(1) - (2) get
(1-x)S =1+x+x^2+x^3…… x^(n-1)-nx^n
=(1-x^n)/(1-x)-nx^n
S=(1-x^n)/(1-x)^2-nx^n/(1-x)

Odd times even = {} odd times odd = {} odd plus even = {} even times even = {} odd times odd = {}

Odd multiplied by even = {odd} odd multiplied by odd = {even} odd plus even = {odd} even multiplied by even = {even} odd multiplied by odd = {odd}

Is odd multiplied by odd equal to even

Odd times odd equals odd

Even sums within 100 are______ ．

Even numbers within 100 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, a total of 50, 2 + 4 + 6 + 8 + +92 + 94 + 96 + 98 + 100 = (2 + 100) × 50 / 2 = 102 × 50 / 2 = 2550 answer: the sum of even numbers within 100 is 2550

When n minimum continuous positive even numbers are added, what is the relationship between their sum and s, N, can be expressed by formula
Sum of addends m (s)
1 2=1×2
2 2+4=6=2×3

n=1 S1=2 = 1×2
n=2 S2=2+4 =2×3
n=3 S2=2+4+6=3×4
……
n=n Sn=2+4+…… +2n=n(n+1)

It is known that the sum of odd terms of arithmetic sequence an is 51, the sum of even terms is 45.5, the sum of even terms is 1, and the number of terms is odd?
It is known that the sum of odd terms of an is 51, the sum of even terms is 45.5, the sum of prime terms is 1, and the number of terms is odd?

The data you gave is wrong
If n is set, then the middle term is the (n + 1) / 2 term, and it is set as a
The sum of odd terms is equivalent to ((n + 1) / 2) * a = 51
The sum of even terms is equivalent to ((n-1) / 2) * a = 45.5. In this way, we can calculate n and a, but the calculated n is a decimal because the value you gave is wrong
Then calculate the general term: a = 1 + ((n + 1) / 2-1) * d = A. A and n have been solved in the previous step, so D is solved
The general term formula is 1 + (k-1) d
The last term is 1 + (n-1) * D
complete.

The sum of odd terms of arithmetic sequence {an} is 51. The sum of even terms is 22.5. The first term is 1. The number of terms is odd. Find the formula of last term and general term

Let an = 1 + (n-1) * K
A(n-1)=1+(N-2)*K
A1=1
A2=1+K
51= (（A1+An)*(N+1)/2 )/2
22.5=( (A2+A(n-1))*(N-1）/2 )/2
Work out n and K ~ and you'll get it

Let the sum of odd and even terms be 44 and 33, respectively

Let {an} be 2n + 1, s odd = a1 + a3 + +A2N + 1 = (n + 1) (a1 + A2N + 1) 2 = (n + 1) an + 1, s-even = A2 + A4 + A6 + +A2N = n (A2 + A2N) 2 = Nan + 1, s odd s even = n + 1n = 4433, the solution is n = 3, 2n + 1 = 7, because s odd s even = an + 1 = a, so A4 = s odd s even = 44-33 = 11, so the middle term is 11

Note CN = an × BN, find the first n terms and TN of sequence {CN}, online and so on! an=4n-11 bn=2^n-1