1 2 3 2010 ？How to answer

It's all the same formula:

First + Last)*Number of Items 2

It's just that the last 4 changes are small.

The principle is the same: such as 1+3+5+......2009=(1+2009)*1005/2

The number of items and the number of items are just off.

The last one takes a 2 in front of it and becomes: 2+4+6+......2n=2(1+2+3+..n) Same as the first one:

The answers are: (1+n)*n 2

1+2n-1)*n/2=n^2

2+2n)*n/2

You follow the formula!

Let me derive the formula for summing the series of equal differences:

s=a1 + a1+d)+ a1+2d)+…a1+(n-1)d](1)

s=[a1+(n-1)d]+(a1+(n-2)d]+…a1(2)2s=(2a1+(n-1)d)+(2a1+(n-1)d)+…2a1+(n-1)d)

2s=[2a1+(n-1)d]*n

s=a1*n+n(n-1)d/2

where a1 represents the first item you requested, i.e., the first number.

d represents your tolerance, which is the difference between the second number and the first.

Equations for the difference series.

First + Last)*Number of Items 2

Haven't you learned about proportional functions? Certainly didn't go to high school.

First year of high school, Chapter 3, Number Sequence Problems

This problem requires the rock block to use the formula of summing the number series from the finch to the equal difference:

Original Rough Years Hu Style = [(Prime Minister Ten Late Phases) Number of Items] 2

1+2014)x2014÷2

So 1+2+3+4+.2014=2029105<>

Adopt the code Li Gao Trap Modulus Trillion Wang Rent algorithm:

sum=1+2+3+4+…Chain line Kai +2013+2014

sum=2014+2013+…+4+3+2+1

2×sum=2015+2015+2015+…+2015+2015, a total of 2014 with towns 2015 added, so it is 2014*2015 shed call 2

Solution: 1+2+3+......2010

Additional Answer: For a set of numbers that are equally different, sum :

AND = (1st digit + last digit) The total number of digits is 2

Because 1 2 3 ......n=n(n 1) 2 So2010x (1 2 3 ......n)=2010x both sides divided by 2009 to get 4020x 2022=1 So the solution of the equation is x=1011 2010

In the future, like this question type, you will directly add the head and tail, multiply by the number, and finally divide by 2, and the final result is equal to 2023066. Thank you.

2023066, just sum according to the equation of equal difference, (first term + last term) x term number 2

Sum of the difference series: (1 2011) 2011 2 2023066

1 All the way up to a hundred.

Let s=1 +2 +3 +....2011, then s=2011+2010+2009+......1 Add the two formulas, 2s = (1 + 2011) + (2 + 2010) + (3 + 2009) + ....2011+1）

That is, there are 2011 groups on the right (there are only 2011 numbers in each of the original two formulas), and the sum of each group is 2012, so 2s=2011*2012

Therefore, s=2011*1006=20230661,2,3,......2011 is a series of equal differences with 1 as the first term and tolerance, that is, a1=1, d=1, so s2011 (the sum of the previous 2011 terms) = 2011*a1+2011*(2011-1) 2*d=2023066

(1+2011)x2011 2 The first term plus the last term Multiply the number of terms The total is divided by 2 The answer is fine.