A problem of summation of sequence It is known that an = (2n + 1) 2n Find SN

an=（2n+1）2n =4n^2+2n
Sn=2n(n+1)(2n+1)/3+n(n+1)

A problem of summation of sequence
1/2n+3/4n+5/8n+...+（2n-1）/n*2^n

The original formula / 2 = 1 / 4N + 3 / 8N +... + (2n-1) / N * 2 ^ (n + 1), so the original formula / 2 = 1 / 2n + 2 / 4N + 2 / 8N +... + 2 / N * 2 ^ n - (2n-1) / n * 2 ^ (n + 1) n * original formula / 2 = 1 / 2 + 1 - (1 / 2) ^ (n-1) / 2 ^ (n + 1) original formula = 3 / N - (2n + 7) / 2 ^ (n + 1)

A problem about summation of sequence
1/1×3 + 1/3×5 +1/5×7 +.+ 1/99×101=

Methods: the method of elimination by separating items! For example, 1 / 1 × 3 = 1 / 2 (1 / 1-1 / 3)
1/3×5 =1/2（1/3-1/5）
So 1 / 1 × 3 + 1 / 3 × 5 + 1 / 5 × 7 +. + 1 / 99 × 101 = 1 / 2 (1 / 1-1 / 3) + 1 / 2 (1 / 3-1 / 5) +. + 1 / 2 (1 / 99-1 / 101) = 1 / 2 [1 / 1-1 / 3 + 1 / 3-1 / 5 +. + 1 / 97-1 / 99 + 1 / 99-1 / 101] = 1 / 2 [1 / 1-1 / 101] = 50 / 101!
Comments: This is a kind of sequence summation problem! The method used is the elimination method!

Sequence problem, sum
（1）（a-1）+(a^2-2)+…… +(a^n-n)
(2) 1+2x+3x^2+…… nx^(n-1)
Please write down the specific steps

(1)
（a-1）+(a^2-2)+…… +(a^n-n)
=(a+a^2+… +a^n)-(1+2+… +n)
1+2+… +n=n(n+1)/2
If a = 1, then a + A ^ 2 + +a^n=n
Then the original formula = N-N (n + 1) / 2 = n (1-N) / 2
If a ≠ 1, then a + A ^ 2 + +a^n=a*(1-a^n)/(1-a)
Then the original formula = a * (1-A ^ n) / (1-A) - n (n + 1) / 2
(2)
Let s = 1 + 2x + 3x ^ 2 + +nx^(n-1) ①
When x = 1, s = 1 + 2x + 3x ^ 2 + +nx^(n-1）
=1+2+3+… +n
=n(1+n)/2
If x ≠ 1
xS=x+2x^2+3x^3+… +(n-1)x^(n-1)+nx^n ②
① (1-x) s = 1 + X + x ^ 2 + +x^(n-1)-nx^n
=(1-x^n)/(1-x)-nx^n
Then s = (1-x ^ n) / (1-x) ^ 2-nx ^ n / (1-x)

Known sequence an = 1 / [n (n + 1) (n + 2)]
Sum SN
(i.e. a1 + A2 + a3 + A4 +... + an)

an=1/[n(n+1)(n+2)]=1/2[1/n(n+1)-1/(n+1)(n+2)]
Sn=1/1*2*3+1/2*3*4+.+1/[n(n+1)(n+2)]
=1/2[1/1*2-1/2*3]+1/2[1/2*3-1/3*4]+...+1/2[1/n(n+1)-1/(n+1)(n+2)]
=1/2[1/1*2-1/2*3+1/2*3-1/3*4+...+1/n(n+1)-1/(n+1)(n+2)]
=1/2[1/1*2-1/(n+1)(n+2)]
=1/2*[(n+1)(n+2)-2]/[2(n+1)(n+2)]
=(n^2+3n)/[4(n+1)(n+2)]

What is the sum of arithmetic sequence in primary school 1 + 3 + 4 + 7 + 9 + 10 + 12 + 13 +. + 66 + 67 + 69 + 70

Divide into arithmetic sequence
1+4+7+10+13+...+67+70=（1+70）/2*24=852
3+6+9+...+66+69=（3+69）/2*23=828
So the original formula = 852 + 828 = 1680

In the sequence {an}, A1 = 2, an = 2An − 1 + 2n + 1 (n ≥ 2, n ∈ n *) (1) let BN = an2n, prove that {BN} is an arithmetic sequence; (2) under the condition of (1), let TN = 1b1b1b2 + 1b2b3 + +1bnbn + 1, find TN

(1) It is proved that from an = 2An − 1 + 2n + 1, an2n = an − 12n − 1 + 2 (4) an2n − an − 12n − 1 = 2 (n ≥ 2) (5 points) BN = an2n, ∧ B1 = 1, ∧ sequence {BN} is an arithmetic sequence with the first term of 1 and the tolerance of 2 (6 points) (2) from (1), we know that BN = 2N-1, 1bnbn + 1 = 1 (2n − 1) (2n + 1) = 12 (12n − 1 − 12n + 1) (9) TN = 12 (1 − 13 + 13 − 15 + +12n−1−12n+1)=12(1−12n+1)=n2n+1… (12 points)

Sum of arithmetic sequence 10 questions,

The expression of 2 + 4 + 6 + 8 +... + 2n summation uses n to indicate how to push down
1 + 2 + 3 + 4 +... + N + (n + 1) + (n + 2) +... 2n sum
1 + 3 + 5 + 7 +.. + (2n-1) how to push down the summation expression with n
The first (x + 1) ^ 2, the third (x-1) ^ 2, the sum of the first 13 items is 520, find the value of X
The sum formula of the first n terms of a sequence is Sn = 5 ^ n + 1, which proves that it is an equal ratio sequence
Summation of sequence
(1) (K-2) * (K + 3) K from 1 to n
(2) ((K-2) * (K + 3)) ^ (- 1) K from 1 to n
(3) (2k + 1) / (k ^ 2 * ((K + 1) ^ 2)) K from 1 to n

Summation of higher one sequence
1× 1/2 + 2×1/4 × 3 × 1/8 × …… The nth power of x n × 1 / 2
>

Sn=1*(1/2)+2*(1/2^2)+3*(1/2^3)+...+n(1/2^n)
Sn/2=1*(1/2^2)+2*(1/2^3)+3*(1/2^4)+...+n[1/2^(n+1)]
Up down
Sn/2=1/2+1/2^2+1/2^3+...+1/2^n-n[1/2^(n+1)]
=(1/2)(1-1/2^n)/(1-1/2)-n[1/2^(n+1)]
=(1-1/2^n)-n[1/2^(n+1)]
So Sn = 2 (1-1 / 2 ^ n) - 2n [1 / 2 ^ (n + 1)]
This is a commonly used recursive sequence of dislocation subtraction!

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．

A problem of summation of sequence It is known that an = (2n + 1) 2n Find SN