Numerical reasoning How to solve these, numerical reasoning solving

Question 1: Select B

For the sake of writing convenience, set 3 values first:

k=(-1)^[n(n+1)/2]

a=(1-k)/2

b=(1+k)/2

General formula: 90a+863b+(-1) n (73a+648b) Do the math yourself! That's really hard for you!! I can't stand it, is it a civil service exam?

I also found this answer from the Internet.

If I were to choose 189, it would start with 1!! The second question consists of a sequence of 6, 4, 10, in which the 3rd term = the sum of the first two terms, so the sum of the numbers in the choice option = 14.

This is D 77 Question 3.

So the answer is 145

Tired!!! I'll take a look at the next question!! Question 4.

So you know it, right?

Question 5. Can't find + won't!

Question 6. The difference is: 2, 6, 18, 54, 2, 2*3, 6*3, 18*3 then the post-90s should be 90+2*3*3*3*3 I finally know how amazing the network is!!

You ask so many questions that someone actually has you !!

I'll just do it!! I'm convinced!!

I've been trained by yourself on this question!!

Question 5: I will do it!!

Can you see it?? 3= 3 to the power!!

What a cow upstairs! I won't answer them all, but I will only answer a few of them.

…The core of the cave shouted grinding .........

The answer is 100 = 4 25 = 4 (21 + 8 2).

The answer is d 1 4 The first three numbers are 8 6 4, and the difference between the two numbers is 2. From this, it can be seen that the difference between the two before and after the last three numbers should also be the same. 9 4 is 4 4 different from 5 4, so the last number should be 4 4 different from 5 4, i.e. the last number is 1 4

Strictly speaking, there is no pattern.

According to the method of equal difference: 8-2 = 6-2 = 4, 4-9 4 = 7 4. And 9 4-5 4 = 4 4.

With the expansion coefficient method: original, 32 4;24/4；16/4；9/4；5/4 ..The difference in the molecules is , and there is still no regularity to speak of.

Question 1: 11 2, 15, 67 2, (219 2 ?)

a：62b：63

c：64d：65

The answer is that the c denominator is 2

The numerator is a cubic sequence of +3.

So the answer is 128 2 = 64

Question 2: 3, 8, 17, 32, (

a：46b：53

c：57d：61

The answer is c, which is broken down into two sequences.

Add up and down to get the original sequence 25 + 32 = 57

Questions 3: 1, 16, 27, 16, (

a：1b：5c：2

D:3 The answer is B

Respectively.

power. Question 4: 3, 1, 1, 6, 25, (

a：62b：58

c：68d：74

The question is wrong, and the second item should be -2

It is obtained from this sequence -2.

(64) Cubic sequence.

So the answer is 62-2=62 choose A

10 Each two items are divided into a group, i.e. the term is a group of items and the term is a group...

Each group is summed. The sum of the first set of two numbers is 2

The sum of the two numbers in the second group is 4 (2*2).

The sum of the three numbers is 8 (4*2).

It can be seen that the sum of the numbers in each group is equal to 2 times the number of the previous group.

The sum of the four sets of two numbers should be 8*2=16

So the numbers in parentheses are 16-6=10

c can be converted into 10+1, 20+2, 30+3, 40+5, (50+7), 60+11

A series of equal differences before the sign.

The sign is followed by a sequential sequence of non-composite numbers.

These questions are very simple, and the details are as follows:

x3+1=13 13x3+1=40 40x3+1=121 121x3+1=364 The parentheses should be 364x3+1=1093

2. Are you asking the wrong question? I've done this problem, it should be 1,1 3,1 3,1 9,1 27,( I can't answer what you said, if it's 1,1 3,1 3,1 9,1 27,() then it's easy to solve, directly the first two denominators are multiplied to equal the third denominator, and the numerator is all 1 without consideration, and 1 243 is obtained.

3. The parentheses and 23 38 after the third question are reversed, it should be 23 38 in the front, the parentheses in the end, and the variant is 1 1, 2 4, 6 11, 17 29, 46 76 () The numerator of the previous term plus the denominator = the numerator of the latter term, the denominator of the previous term plus the numerator of the latter term + 1 = the denominator of the latter term, and 122 199 is obtained.

First question: It can also be 3 squared plus 1st term = 2nd term, 3 to the 3rd power + 2nd term = 3rd term, 3 to the 4th power + 3rd term = 4th term, 3 to the 5th power + 4th term = 5th term, 3 to the 6th power + 5th term = 1093

Question 2: Multiply the preceding and posterior terms by imaginary 1 81, choose a

Question 3: b can be rewritten as 1 1 2 4 6 11 17 29 23 38 (46 76) 122 199

The numerator plus the denominator of the previous term = the numerator of the latter term, and the denominator of the previous term plus the numerator of the latter term + 1 = the denominator of the latter term.

The first number: The latter number is the previous number multiplied by three plus one.

Second pass: 3 = 3 1

In this case, the denominator Chen San is.

Then the denominator of the first two is multiplied to get the denominator of the third number.

The third way: b is divided first: 1 1, 2 4, 6 11, 17 29, 46 76 The numerator of each fraction is the numerator and denominator of the previous fraction, and the denominator is the sum of the denominator of the previous fraction plus 1