In the sequence an, we know that A1 = - 1, a (n + 1) = Sn + 3n-1 to find an

In the sequence an, we know that A1 = - 1, a (n + 1) = Sn + 3n-1 to find an

Sn = a (n + 1) - 3N + 1sn + 1-sn = a (n + 2) - A (n + 1) - 3 = a (n + 1) a (n + 2) = 2A (n + 1) + 3 let a (n + 2) + λ = 2 [a (n + 1) + λ] then a (n + 2) = 2A (n + 1) + λ get λ = 3, so a (n + 2) + 3 = 2 [a (n + 1) + 3] so {an + 3} is an equal ratio sequence

There are 21 matches on the table. Xiaogang and Xiaoliang take turns to take one or two matches each time. Whoever gets the last one wins. How can Xiaoliang win?

According to the above analysis, as long as the total number of sticks is divided by the sum of the sticks taken by two people each time, if there is no remainder, let the other party take it first. Xiaoliang lets Xiaogang take it first, and the total number of sticks taken by himself and Xiaoming is 3, then Xiaoliang guarantees to win

The side expansion of a cylinder is a square with a side length of 9.42 decimeters. How many square decimeters is the side area of the cylinder?

9.42 × 9.42 = 88.7364 (square decimeter)

How many days does the earth go round the sun
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A year 365 days 4 hours 58 minutes 56 seconds
So there will be a leap year of 366 days in four years

Calculation of physical specific heat capacity,
When sawing wood with a saw, 20 times per minute, moving 0.8m each time, the force of human sawing is 150n. If 70% of the work done during sawing is consumed in the heat generated, how much heat is generated in 10min?

Q=W=10*20*150*0.8*70%=16800J

There is a cylinder. The circumference of its bottom surface is equal to its height. If the height is shortened by 2 cm, the surface area will be reduced by 6.28 cm. Find the volume of the cylinder

Bottom radius: 6.28/2/3.14/2 = 0.5
Height: 3.14 * 0.5 * 2 = 3.14
3.14*0.5*0.5*3.14=2.4649

1 / 3 to 5 / 18

One in three to five in 18
=6：5

It is proved that: (1) Tan & sup2; α - Sin & sup2; α = Tan & sup2; α × Sin & sup2; α (2) (COS β - 1) & sup2; sin & sup2; β = 2-2cos β (3)

(1) It is proved that: Tan & sup2; α - Sin & sup2; α = Sin & sup2; α / cos & sup2; α - Sin & sup2; α = Sin & sup2; α (1 / cos & sup2; α - 1) = Sin & sup2; α (1-cos & sup2; α) / cos & sup2; α = Sin & sup2; α * Sin & sup2; α / cos & sup2; α = Tan & sup2; α × Sin & sup2; α (2)

As shown in the figure, a road with the same width shall be built on the rectangular ground with a width of 20m and a length of 30m, and the remaining part of the road shall be planted with lawn with an area of 551m & sup2; to calculate the width of the road

Let the width of the road be XM
（30－x）（20－x）＝551
x²－50x＋49＝0
（x－1）（x－49）＝0
X1 = 1, X2 = 49

3 / 4 to 16, 2.5 to 75, 8 / 3 to 1.25, 8 / 5 to 1, 5 to 1
There are still 7 / 20 to 4 / 5 ratio, 5 / 6 ratio to 1500 kg ratio, 2 / 3 and 3 / 4 ratio, faster than 34 / 3 ratio, please

3/4：16=3：64=3/64
2.5：75=5：150=1：30=1/30
8/3：1.25=64：30=32/15
8/5：1.5=8/5：3/2=16：15=16/15
7/20：4/5=7：16=7/16
5 / 6 ton: 1500 kg = 5 / 6 ton: 1.5 ton = 5 / 6:3 / 2 = 5:95 / 9
（2/3+3/4）：4/3=17/12：4/3=17：16=17/16