1-2²+3²-4²+5²-6²+.+99²-100²

1-2²+3²-4²+5²-6²+.+99²-100²

simple form
=(1+2)(1-2)+(3+4)(3-4)+...+(99+100)(99-100)
=-(1+2)-(3+4)-(5+6)-...-(99+100)
=-(1+2+3+… +99+100)
=-5050

Sum: SN = 1 / 2 ^ 2-1 + 1 / 4 ^ 2-1 +. 1 / (2n) ^ 2-1

If the nth addend is: 1 / [(2n) & # - 1] = 1 / [(2n + 1) (2n-1)] = (1 / 2) [1 / (2n-1)] - [1 / (2n + 1)], then:
S=(1/2){[(1/1)－(1/3)]＋[(1/3)－(1/5)]＋… ＋[1/(2n－1)－1/(2n＋1)]}
=(1/2)[1－1/(2n＋1)]
=(n)/(2n＋1)

Sum 1 / 1,1 / 2,2 / 2,1 / 3,2 / 3,3 / 3,1 / 4,2 / 4,3 / 4,4 / 4 1/100,2/100,3/100,…… 99/100,100/100.
According to a certain rule, there are a series of numbers: 1 / 1, 1 / 2, 2 / 2, 1 / 3, 2 / 3, 3 / 3, 1 / 4, 2 / 4, 3 / 4, 4 / 4 1/100,2/100,3/100,…… 99 / 100 / 100. What's the sum of these numbers?

If the denominator is n, it is divided into one term, and the sum of this term is n * (n + 1) / 2 / N = 1 / 2 + n / 2
If there are 100 such items, Hewei 1 / 2 * 100 + (1 + 2 +... + 100) / 2 = 2575;

Sequence: 1,1 / 2,1 / 3,1 / 4,1 / 5., sum
World class problem!

Hello, 1,1 / 2,1 / 3.1/n is called harmonic series in mathematics
The sum formula of the first n terms does not exist,
When n tends to infinity, the limit of 1 + 1 / 2 + 1 / 3 + 1 / 4 +. + 1 / N is infinity
But the value of 1 + 1 / 2 + 1 / 3 + 1 / 4 +. + 1 / n-ln (n) tends to a constant when n tends to infinity, which is called Euler gamma, or Euler constant, about 0.577216;
And 1 / N + 1 / (n + 1) + 1 / (n + 2) +. 1 / (2 * n) when n tends to infinity, the sum is ln (2);
If you are interested in further understanding, you can check these: Euler constant, harmonic series, Bernoulli natural number power sum formula, Riemannian function (zeta function), Bernoulli series, Euler formula (that is, the formula for even number of Riemannian function variables)

Sequence summation 1 / 1 * 3,1 / 2 * 4,1 / 3 * 5... 1 / N (n + 2)
RT

1 / 1 * 3 = (1 / 2) (1-1 / 3) 1 / 2 * 4 = (1 / 2) (1 / 2-1 / 4)... 1 / N (n + 2) = (1 / N - 1 / (n + 2)) / 2 original formula = (1 / 2) (1-1 / 3 + 1 / 2-1 / 4 +... + 1 / (n-1) - 1 / (n + 1) + 1 / (n) - 1 / (n + 2)) = (1 / 2) (1 + 1 / 2-1 / (n + 1) - 1 / (n + 2)) = 3 / 4 - (2n + 3) / 2 (n + 1) (n + 2)

Sequence summation 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5. + 1 / n

Haha, it's not harmonic series. Limit divergence. As for the sum of the first n terms, it's more complicated

Summation of sequence 2 ^ 2 + 4 ^ 2 + 6 ^ 2 + 8 ^ 2 + 10 ^ +

The original formula = 2 * (1 + 2 + 3 + 4 +...) = 2 * 1 / 6 * n * (n + 1) * (2n + 1) = 2 / 3 * n * (n + 1) * (2n + 1)

Sum of sequence 3 * 2 + 6 * 4 + 9 * 8 +... + 3N * 2 ^ n

S=3*2+6*4+9*8+...+3n*2^n
2S=3*4+6*8+…… +3(n-1)*2^n+3n*2^(n+1)
-S=3*2+3*4+3*8+…… +3*2^n-3n*2^(n+1)
S=3n*2^(n+1)-6*[(1-2^(n-1))/(1-2)]
=3n*2^(n+1)-6*2^(n-1)+6

What are the factors from 1 to 30

Supplement by the respondent 2009-09-05 08:14
Do you want to factor all these numbers?
1 1
2 1 2
3 1 3
4 1 2 4
5 1 5
6 1 2 3 6
7 1 7
8 1 2 4 8
9 1 3 9
10 1 2 5 10
11 1 11
12 1 2 3 4 6 12
13 1 13
14 1 2 7 14
15 1 3 5 15
16 1 2 4 8 16
17 1 17
18 1 2 3 6 9 18
19 1 19
20 1 2 4 5 10 20
21 1 3 7 21
22 1 2 11 22
23 1 23
24 1 2 3 4 6 8 12 24
25 1 5 25
26 1 2 13 26
27 1 3 9 27
28 1 2 4 7 14 28
29 1 29
30 1 2 3 5 6 10 15 30

What are the factors of 30

1,2,3,5,6,10,15,30