Sum: 1 + 3 + 5 + ＋(2n－1)

Sum: 1 + 3 + 5 + ＋(2n－1)

The difference is equal to the square of the answer n. just set up a formula

Sum 1 + 3 * 3 + 5 * 3 ^ 2 +... + (2n-1) * 3 ^ (n-1)

Sn = 1 + 3 * 3 + 5 * 3 ^ 2 +... + (2n-1) * 3 ^ (n-1)
3Sn=1*3+3*3^2+...+(2n-1)*3^n
Dislocation subtraction
-2Sn=1+2*3+2*3^2+2*3^3+...+2*3^(n-1)-(2n-1)*3^n
=2(3^n-n*3^n-1)
Sn=(n-1)*3^n+1

Sum Sn = 1 + (1 + 3) + (1 + 3 + 3 ^ 2) + (1 + 3 + 3 ^ 2 + 3 ^ 3) +. + (1 + 3 + 3 ^ 2 + 3 ^ 3 +... + 3 ^ n-1)

A1 = 3 ^ 0, A2 = 3 ^ 0 + 3 ^ 1, A2 = 3 ^ 0 + 3 ^ 1 + 3 ^ 2, so an = 3 ^ 0 + +3 ^ (n-1), with n terms, q = 3, so an = 3 ^ 0 * (3 ^ n-1) / (3-1) = (1 / 2) * 3 ^ n-1 / 2, so Sn = [(1 / 2) * 3 ^ 1-1 / 2] + [(1 / 2) * 3 ^ 2-1 / 2] + +[(1/2)*3^n-1/2] =(1/2)*(3^1+3^2+…… +3^n)-1/2*n =(1/2)*3*(3^n-1)/(3-1)-n/2 =(3/4)*(3^n-1)-n/2

Sum Sn = 1 × 2 + 2 × ^ 2 + 3 × 2 ^ 3 +... + n × 2 ^ n

Sum: SN = 1 + (1 + a) + (1 + A + A ^ 2) +. + (1 + A + A ^ 2. + A ^ n)
How to find the sum of this sequence?

Sn*(1-a)=(1-a)+(1-a^2)+(1-a^3)+.+(1-a^(n+1))
Sn*(1-a)=(n+1)-(a+a^2+...+a^(n+1))
After that, you don't have to teach
The key is the first step, multiply both sides by (1-A)

The first n terms of {an} and Sn = 1-5 + 9-13 + 17-21 + are known +(- 1) n + 1 (4n-3), then the value of s22-s11 is______ ．

According to the meaning of the title, it is easy to get S22 = 1-5 + 9-13 + 17-21 + +81-85=（1-5）+（9-13）+（17-21）+… +（81-85）=（-4）×11=-44，S11=1-5+9-13+17-21+… +33-37+41=（1-5）+（9-13）+（17-21）+… +(33-37) + 41 = (- 4) × 5 + 41 = 21, then s22-s11 = - 44-21 = - 65

Sum: SN = - 1 - (5 / 9) - (7 / 27) +. + (- 2n-1) (1 / 3) ^ n

Sn = - 1 - (5 / 9) - (7 / 27) +. + (- 2n + 1) (1 / 3) ^ (n-1) + (- 2n-1) (1 / 3) ^ n3sn = (- 3) * (1 / 3) ^ 0 + (- 5) * (1 / 3) ^ 1 +. + (- 2n-1) (1 / 3) ^ (n-1) subtraction: 2Sn = (- 3) * (1 / 3) ^ 0 + (- 2) [(1 / 3) ^ 1 + (1 / 3) ^ 2 + (1 / 3) ^ 3 +. + (1 / 3)

Sum: SN = 1 + (1 + 12) + (1 + 12 + 14) + [1 + 12 + 14 + +（12）n-1]．

∵1+12+14+… +（12）n-1=1−(12)n1−12=2−12n−1，∴Sn＝2n−(1+12+122+… +12n−1)=2n-1−12n1−12=2n-2+12n−1．

Sum Sn = 1 × 2 × 3 + 2 × 3 × 4 + +n(n+1)(n+2)
Five stars

Using combination number formula
n(n+1)(n+2)=6*C(n+2,3)
Sn=6[C(3,3)+C(4,3)+C(5,3)+.+C(n+2,3)]
Sn=6[C(4,4)+C(4,3)+C(5,3)+.+C(n+2,3)]
Continuous use of the formula C (n, m) + C (n, m-1) = C (n + 1, m)
Sn=6[C(5,4)+C(5,3)+.+C(n+2,3)]
=6[C(6,4)+.+C(n+2,3)]
=6C(n+3,4)
=(n+3)(n+2)(n+1)n*6/24
=(n+3)(n+2)(n+1)n/4

What is the summation formula of 1 + 2 + 4 + 8 + 16 + 32?
Sum of equal ratio sequence
Sn=a1(1-q^n)/(1-q)=(a1-anq)/(1-q)
What I found is this, but I don't know what s is and what n is
And what's a, what's Q, what's A1
By the way, do a question for me, 1 + 2 + 4 + 8 + 16 + 32... 16384 =?

If the ratio of each term to the previous term of a sequence from the second term is equal to the same constant, the sequence is called equal ratio sequence. This constant is called the common ratio of equal ratio sequence, which is usually represented by the letter Q (Q ≠ 0)
(1) The general formula of equal ratio sequence is: an = A1 * q ^ (n-1)
If the general term formula is transformed to an = A1 / Q * q ^ n (n ∈ n *), then an can be regarded as a function of the independent variable n when Q > 0, and the points (n, an) are a group of isolated points on the curve y = A1 / Q * q ^ X
(2) Sum formula: SN = Na1 (q = 1)
Sn=A1(1-q^n)/(1-q)
=(a1-a1q^n)/(1-q)
=A1 / (1-Q) - A1 / (1-Q) * q ^ n (i.e. a-aq ^ n)
(premise: q is not equal to 1)
The relation between any two terms am and an is an = am · Q ^ (n-m)