Sum, 1 / 4-1 / 8 + 1 / 16-1 / 32 +. - 1 / 1024 emergency

Sum, 1 / 4-1 / 8 + 1 / 16-1 / 32 +. - 1 / 1024 emergency

Sum of equal ratio sequence
Sn=a1(1-q^n)/(1-q) =(a1-an×q)/(1-q) (q≠1)
The common ratio is (- 1 / 2)
Sn=（1/4）*(1-(-1/2)^99)/(1-（-1/2）)

Sum: 1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 + 1 / 32 + 1 / 64 +. + 1 / 1024 =?

This is an equal ratio sequence with 1 / 2 as the common ratio
1/2+1/4+1/8+1/16+1/32+1/64+.+1/1024
=1/2+1/4+1/8+1/16+1/32+1/64+.+1/2^10
=(1/2)(1-(1/2)^10)/(1/2)=1023/1024

-Sum of 1,2, - 4,8, - 16,32,..., 512

1+2+4+8+16+32+64+128+256+512+1024
=2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10
Let s = 2 ^ 0 + 2 ^ 1 + 2 ^ 2 + 2 ^ 3 + 2 ^ 4 + 2 ^ 5 + 2 ^ 6 + 2 ^ 7 + 2 ^ 8 + 2 ^ 9 + 2 ^ 10
2s=2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10+2^11
2s-s=2^11-2^0
s=2^11-1
s=2048-1
s=2047
That is, 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 = 2047

Question: let the general term formula of sequence {an} be an = 1 / N2 + 4N + 3, then what is the sum of the first n terms?
5/12-1/2n+4-1/2n+6

an=1/(n+1)(n+3)=1/2*[1/(n+1)-1/(n+3)]
So Sn = 1 / 2 * [1 / 2-1 / 4 + 1 / 3-1 / 5 + +1/n-1/(n+2)+1/(n+1)-1/(n+3)]
=1/2*[1/2+1/3-1/(n+2)-1/(n+3)]
=(5n^2+13n)/(2n^2+10n+12)

Sum of sequence exercises
I am very weak in this area. I hope you can provide some questions with answers

(Lecture 13) general term formula of sequence and common methods of summation

Ask the question of summation of sequence
How to sum 1 + 1 / 2 + 1 / 3 + 1 / 4 +. + 1 / N?

1+1/2+1/3+… +There is no good calculation formula for 1 / N, all calculation formulas are approximate values, and the accuracy is not high. The sequence composed of the reciprocal of natural numbers is called harmonic sequence. People have studied it for hundreds of years. But so far, there is no summation formula, only the approximate formula

Summation of series
Is there any other way to find the sum function of series ∑ x ^ n / N! Besides expanding with e ^ x? For example, I didn't know that e ^ x could be expanded like that, so I will use other methods. Thank you

f(x) = ∑x^n/n!
∫f(x) dx = ∫∑x^n/n!dx
∫f(x) dx = ∑∫x^n/n!dx = ∑ x^(n+1)/(n+1)!= f(x) - 1
f(x) = e^x+C
F (0) = 1 (Note: 0 ^ 0 / 0! = 1)
f(x) = e^x

1^2+2^2+…… +n^2=?

1^2+2^2+3^2+…… +N ^ 2 = n (n + 1) (2n + 1) / 6 using the cubic difference formula n ^ 3 - (n-1) ^ 3 = 1 * [n ^ 2 + (n-1) ^ 2 + n (n-1)] = n ^ 2 + (n-1) ^ 2 + n ^ 2-N = 2 * n ^ 2 + (n-1) ^ 2-N ^ 2 ^ 3-1 ^ 3 = 2 * 2 ^ 2 + 1 ^ 2-2 3 ^ 3-2 ^ 3 = 2 * 3 ^ 2 + 2 ^ 2 ^ 2-2-2-2 ^ 2 + 2 ^ 2 ^ 2 ^ 3-3 = 2 * 4 ^ 2 + 3 ^ 2-4

Sum: SN = 12 + 34 + 58 + 716 + +2n−12n．

Because Sn = 12 + 34 + 58 + 716 + +2n − 12n, so 12sn = 14 + 38 + 516 + +2n − 32n + 2n − 12n + 1, the subtraction of the two formulas is: 12sn = 12 + 24 + 28 + 216 + +22n − 2n − 12n + 1 = 12 + 12 (1 − 12n − 1) 1 − 12 − 2n − 12n + 1, then Sn = 3 − 2n + 32n

Sn=1x2/2 + 2x3/2 +3x4/2+.+nx(n+1)/2 = __________

You put forward 1 / 2, 1x2 + 2x3 +... + n (n + 1) = n (n + 1) (2n + 1) / 6 + n (n + 1) / 2, n (n + 1) = n ^ 2 + N, and then open it, n ^ 2 add n (n + 1) (2n + 1) / 6, n add n (n + 1) / 2, then divide n (n + 1) (2n + 1) / 6 + n (n + 1) / 2 by 2