0,0 ,6 ,24,60 ,120

0,0 ,6 ,24,60 ,120

0.6.24.60.120（210）
law:
0=0*1*2
6=1*2*3
24=2*3*4
60=3*4*5
120=4*5*6
Item n = (n-1) * n * (n + 1)

0,6,24,60,120,210,()
How to do it

The rule is: the last digit minus the first digit = (1 + n) * 6
6-0=1*6=6
24-6=3*6=（1+2）*6=18
60-24=6*6=（1+2+3）*6=36
120-60=（1+2+3+4）*6=60
210-120=（1+2+3+4+5）*6＝90
X-120=(1+2+3+4+5+6)*6=126
X=246
So fill in 246 in brackets

What's the next number of 0, 6, 24, 60120210? What's the rule?
Such as the title

336
The analysis is as follows:
The difference between the numbers is large, but the difference is not too big, we should consider the power, their law is 1 ^ 3-1 = 0, 2 ^ 3-2 = 6, 3 ^ 3-3 = 24, 4 ^ 3-4 = 60, 5 ^ 3-5 = 120, 6 ^ 3-6 = 210
7^3-7=336
^3 is a cubic, or cubic

What's the order of the numbers 0 6 24 60 120 210?

The rule is as follows
0=0*1*2
6=1*2*3
24=2*3*4
60=3*4*5
120=4*5*6
210=5*6*7
Item n = (n-1) * n * (n + 1)

Number reasoning: 0,6,24,60, ()
A125,B89,C126,D120

Dear kandy2001
Choose D, 6-0 = 1 * 6 = 6 24-6 = 3 * 6 = (1 + 2) * 6 = 18 60-24 = 6 * 6 = (1 + 2 + 3) = 36 X-60 = (1 + 2 + 3 + 4) * 6 = 60
60+60=120

1. 1 / 48 + 1 / 80 + 1 / 120 + 1 / 168, 1 / 210 + 1 / 336 + 1 / 504

1 / 48 + 1 / 80 + 1 / 120 + 1 / 168 = 1 / 6x8 + 1 / 8x10 + 1 / 10x12 + 1 / 12x14 = 1 / 2x (1 / 6-1 / 8 + 1 / 8-1 / 10 + 1 / 10-1 / 12 + 1 / 12-1 / 14) = 1 / 2x (1 / 6-1 / 14) = 1 / 2x2 / 21 = 1 / 21210 / 1 + 336 / 210 / 1 + 504 / 1 = 1 / 21x10 + 1 / 21x16 + 1 / 21x24 = 1 / 21x (1 / 10 + 1 / 16

6 24 60 120 ?
According to the first four numbers, we can get the fifth number! Please tell us the answer and give the reason, how to get it!

210
————————————
Analysis——
6+（6*3）=24
24+（6*6）=60
60+（6*10）=120
therefore
120+（6*15）=210

1/6+1/24+1/60+1/120+----+1/117600

Let's look at the numerator 6 = 1 * 2 * 324 = 2 * 3 * 460 = 3 * 4 * 5120 = 4) 5) 6.117600 = 48 * 49 * 50, the denominator is the multiplication of three consecutive natural numbers, and the first natural number of each term plus 1 denominator is = n (n + 1) (n + 2) 1 / 1 * 2 * 3 + 1 / 2 * 3 * 4 + 1 / 3 * 4 * 5 +. + 1 / N (n + 1) (n + 2) +. Sn = 1 / 2 * [1 / 1 * 2-1 / 2 * 3 + 1 /

1 / 6,1 / 24,1 / 60,1 / 120, --, find the general term, and find the sum of the first n terms

General term an = 1 / [n (n + 1) (n + 2)]
an = 1/n * [1/(n+1) - 1/(n+2)]
= 1/2 * [1/n - 1/(n+1) - 1/(n+1) + 1/(n+2)]
The first n terms and Sn = 1 / 2 * [(1 - 1 / 2 - 1 / 2 + 1 / 3) + (1 / 2 - 1 / 3 - 1 / 3 + 1 / 4) + (1 / 3 - 1 / 4 - 1 / 4 + 1 / 5) +... + (1 / (n-1) - 1 / N - 1 / N + 1 / (n + 1)) + (1 / N - 1 / (n + 1) - 1 / (n + 1) + 1 / (n + 2))]
= 1/2 * [(1 - 1/2 - 1/(n+1) + 1/(n+2)]
= 1/4 - 1/[2(n+1)(n+2)]

32 16 24 8 12 20 () () what should I fill in the next two blanks?

24-16=8 8+12=20
so 20-12=（8） 20+8=（28）
Fill in 8 and 28 in two brackets